The ABC's of Number Theory
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Elkies, Noam D. 2007. The ABC's of number theory. The Harvard College Mathematics Review
1(1): 57-76.

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FACULTY FEATURE ARTICLE

6
The ABC’s of Number Theory
Prof. Noam D. Elkies†
Harvard University
Cambridge, MA 02138
elkies@math.harvard.edu
Abstract
The ABC conjecture is a central open problem in modern number theory, connecting results, techniques
and questions ranging from elementary number theory and algebra to the arithmetic of elliptic curves
to algebraic geometry and even to entire functions of a complex variable. The conjecture asserts that,in a precise sense that we specify later, if A, B, C are relatively prime integers such that A + B = C
then A, B, C cannot all have many repeated prime factors. This expository article outlines some of
the connections between this assertion and more familiar Diophantine questions, following (with theoccasional scenic detour) the historical route from Pythagorean triples via Fermat’s Last Theorem tothe formulation of the ABC conjecture by Masser and Oesterlé. We then state the conjecture andgive a sample of its many consequences and the few very partial results available. Next we recite
Mason’s proof of an analogous assertion for polynomials A(t), B(t), C(t) that implies, among otherthings, that one cannot hope to disprove the ABC conjecture using a polynomial identity such as theone that solves the Diophantine equation x2 + y 2 = z 2 . We conclude by solving a Putnam problem
that predates Mason’s theorem but is solved using the same method, and outlining some further open
questions and fragmentary results beyond the ABC conjecture.‡

6.1 Pythagorean triples: x2 + y 2 = z 2
An ordered triple (x, y, z) of integers is called a Pythagorean triple if and only if it solves the Diophantine equation x2 + y 2 = z 2 ; that is, if and only if |x| and |y| are the lengths of the sides, and|z| the length of the hypotenuse, of a right triangle. (We allow degenerate triangles with a “side” oflength zero.) It is well-known that every such triple is proportional to(x, y, z) = (m2 − n2 , 2mn, m2 + n2 )(6.1)

for some integers m, n. Equivalently (dividing by n2 to obtain polynomials in the single rationalvariable t = m/n), the solution (x, y, z) is proportional to (t2 − 1, 2t, t2 + 1) for some t ∈ Q, or to(1, 0, 1) which arises for “t = ∞” (corresponding to (m, n) = (1, 0)). That is, all Pythagorean triples
are accounted for by the single polynomial identity(t2 − 1)2 + (2t)2 = (t2 + 1).

(6.2)† Noam D. Elkies earned his doctorate in mathematics in 1987 at Harvard, where his advisors where Professors Barry Mazur
and Benedict H. Gross. After three years in Harvard’s Society of Fellows he joined the Mathematics faculty and has remained at
Harvard since. Most of his research is in number theory, usually Diophantine geometry (the combination of algebraic geometryand Diophantine equations) and/or computational number theory. Other interests include some combinatorial mathematics (lattices and codes, incidence geometry, and combinatorial games) and, outside of mathematics, classical music (mostly compositionand piano) and chess (usually chess problems and endgames).
‡ Supported in part by NSF grant DMS-0501029.

57

This classical fact can be profitably approached from many points of view.1 In one familiar approach,
illustrating an important method in algebraic geometry, we first divide by z 2 to obtain the equivalent(x/z)2 + (y/z)2 = 1, so we now seek rational solutions of X 2 + Y 2 = 1, or geometrically a rational
point (a point with both coordinates rational) on the unit circle. Note that two nonzero solutions(x : y : z) in integers yield the same solution (X, Y ) in rationals if and only if they are proportional,
so that by going from x2 + y 2 = z 2 to X 2 + Y 2 = 1 we have automatically identified proportional
Pythagorean triples (corresponding to similar right triangles). The unit vector (1, 0) is an obvious
rational point on the circle. This point yields only a degenerate Pythagorean triple, but we can use it tofind any other rational point (X, Y ) using the straight line through (X, Y ) and (1, 0). The general suchline is Y = −t(X − 1), where the slope −t must be rational if X and Y are. (We choose −t rather
than t for consistency with equation (6.2).) Substituting −t(X − 1) for Y in X 2 + Y 2 = 1 we get thequadratic equation X 2 + t2 (X − 1)2 = 1, one of whose solutions must be X = 1. The other solution
is then the root of
X 2 + t2 (X − 1)2 − 1= (t2 + 1)X − (t2 − 1),
X −1that is, X = (t2 − 1)/(t2 + 1). Then Y = −t(X − 1) = 2t/(t2 + 1), so we have recovered the rational
point corresponding to the solution (t2 − 1, 2t, t2 + 1) of x2 + y 2 = z 2 . See Figure 1, which shows
this construction for t = 2.
This procedure readily generalizes: instead of X 2 + Y 2 − 1 we can use any irreducible polynomial
P (X, Y ) of degree 2, and instead of the initial point (1, 0) we can use any rational solution (X0 , Y0 )of P (X, Y ) = 0; the lines through (X0 , Y0 ) not tangent to the curve P (X, Y ) = 0 at that point thenparametrize all other rational points on the curve. [Try X 2 + Y 2 = 2 and X0 = Y0 = 1. What goes
wrong if we attempt this for P (X, Y ) = X 2 + Y 2 and X0 = Y0 = 0? Note that X 2 + Y 2 is irreducible
over the rationals, but not over C where it factors as (X + iY )(X − iY ).] The technique even works insome settings beyond plane curves of degree 2, including notably degree-3 plane curves with a double
point; see Figure 2 for the example of the double point (0, 0) on the curve (X + Y )3 = XY . In
our special case of X 2 + Y 2 = 1 and (X0 , Y0 ) = (1, 0) we can make yet another connection: if(X, Y ) = (cos θ, sin θ) then our line Y = −t(X − 1) makes an angle of θ/2 with the vertical. Thiscan be seen by elementary plane geometry for 0 < θ < π, starting from the fact that (0, 0), (1, 0)
and (X, Y ) are vertices of an isosceles triangle (this too is shown in Figure 1); in general one must
remember that θ is defined only up to integer multiples of 2π. In any case, this gives t = cot(θ/2),
so our parametrization is equivalent to the trigonometric half-angle formulas that give cot(θ/2) as arational function of (sin θ, cos θ) and vice versa:
cot

sin θ
θ
=
;
2
1 − cos θ

cos θ =

cot2 (θ/2) − 1
2 cot(θ/2), sin θ =.
cot2 (θ/2) + 1
cot2 (θ/2) + 1

(6.3)

These
formulas reappear in integral calculus in the guise ofRthe universal substitution that conR
verts f (sin θ, cos θ) dθ (where f is any rational function) into F (t) dt for some rational function
F ∈ R(t), which can then be expanded in partial fractions to √obtain an elementary antiderivative.
Equivalently this lets us integrate any rational function of√X and 1 − X 2 with respect to X, and thegeneralization to quadratic P (X, Y ) = 0 lets us replace 1 − X 2 by the square root of any quadratic
polynomial.
1 Besides the algebro-geometric method we follow, at least four others come to mind, which suggest various perspectives
on and generalizations of the result. The most elementary may be to begin with the trigonometric identities (6.3), or with anequivalent geometric calculation with isosceles and right triangles. An elementary derivation from unique factorization in Z is
obtained by removing common factors from (x, y, z), switching x, y if necessary to make x odd, and using the factorization
x2 = z 2 − y 2 = (z − y)(z + y) and the fact that gcd(z − y, z + y) = 1 to write z ± y = (m ± n)2 for somecoprime integers m, n. See for instance [IR, p.23, Exercise 12]. Alternatively, factor z 2 = (x + iy)(x − iy) in the ringZ[i] of Gaussian integers, and use unique factorization in Z[i]; this explains why x and y are the real and imaginary parts of(m + in)2 . Finally, for X, Y ∈ Q we have X 2 + Y 2 = 1 if and only if the element X + iY of Q(i) has norm 1, which by
Hilbert’s Theorem 90 is equivalent to X + iY = w/w for some nonzero w ∈ Q(i). Taking t = Re(w)/ Im(w) we recoverX + iY = (t2 − 1 + 2it)/(t2 + 1). See [Ta].

58

Y
6

Y
6

A
A
A (3/5, 4/5)
A

 A

A (1, 0)
- X
θ
A
A
A
θ A
2
A Y = −2(X − 1)
A

Y = 2X(2/27, 4/27)





-X







Figure 2: (X + Y )3 = XY
Figure 1: X 2 + Y 2 = 1
But we have digressed from our main plot, to which we now return by looking at x2 + y 2 = z 2
and the parametrization (6.1) or (6.2) from another point of view. We ask: How many solutions does
the Diophantine equation x2 + y 2 = z 2 have in integer triples (x, y, z)? Our parametrizations provide
infinitely many (x, y, z) even when we identify proportional solutions, but we can still ask how common
these solutions are. To make this vague question more precise, for all N > 0 define C(N ) to be
the number of solutions of x2 + y 2 = z 2 in integers such that x2 , y 2 , z 2 are relatively prime and of
absolute value at most N . (We give the condition on x, y, z in this form because of the way we intend
to generalize it to other Diophantine equations, though of course for x2 + y 2 = z 2 the absolute value
condition is equivalent to the single inequality z 2 ≤ N .) Then the existence of infinitely many nonproportional Pythagorean triples is equivalent to the fact that C(N ) → ∞ as N → ∞, and we ask:
How quickly does C(N ) grow?
Using either of the forms (6.1) and (6.2) of our parametrization of Pythagorean triples we see that
C(N ) should grow as some multiple of N 1/2 . For instance, (6.1) gives points (m, n) in the circle
m2 + n2 ≤ N 1/2 , whose number is asymptotic to the area πN 1/2 of the circle. This is not quite
right because we must count only relatively prime (m, n), and if both m and n are odd then we must
remove a common factor of 2; but each of these corrections changes the asymptotic formula only by aconstant factor. As it happens this factor is 2/(3ζ(2)) = 4/π 2 , making C(N ) ∼ (4/π)N 1/2 . But it isthe exponent 1/2 that concerns us here, and we could have guessed this exponent much more easily as
follows. Let A = x2 , B = y 2 , and C = z 2 . ThenA + B = C,
and the number of solutions of A + B = C in relatively prime integers in [−N, N ] is asymptotically
proportional to N 2 . Of the 2N + 1 integers in [−N, N ], approximately N 1/2 are squares (and all but
one are squares in two different ways, but this will not affect the exponent of N , only the coefficientof that power). So, if we pick A, B, C independently and uniformly at random from the integers in[−N, N ], the probability that all three will be squares is asymptotically proportional to N −3/2 . While
we actually choose A, B, C not at random but subject to A + B = C, it seems a reasonable guess thatthe fraction of such (A, B, C) all of which are squares is still roughly N −3/2 , giving a total of roughly3
N 2− 2 = N 1/2 such triples in that range.
If you think this seems suspiciously easy, you are right: we are only guessing the correct answer(up to a constant factor), not proving it. This kind of heuristic is quite naı̈ve, and can easily fail.
For instance, for the equations x2 + y 2 + z 2 = 0 or x2 + y 2 = 3z 2 we might similarly expect the
number of solutions with all three terms in [−N, N ] to grow at the same N 1/2 rate. But neither ofthese equations has any solution other than the trivial (0, 0, 0): the first obviously so, because the terms
x2 , y 2 , z 2 are all nonnegative; and the second because after removing common factors from (x, y, z)
we get a contradiction mod 3.2 In the other direction, the heuristic might grossly underestimate the
2 In fact these two obstructions are more similar than they might seem: x2 + y 2 + z 2 = 0 has no nontrivial solution in the59number of solutions. Consider for example solutions in relatively prime integers of (x + y)3 = xyz(the homogeneous form of the curve (X + Y )3 = XY shown in Figure 2). We might expect very
few solutions, on the grounds that there are about 8H 3 triples (x, y, z) of integers in [−H, H], andin that range (x + y)3 − xyz can be as large as a multiple of H 3 , so should vanish with probability
only c/H 3 for some c > 0, leaving a constant expected number of solutions no matter how large H
is. Somewhat more reasonably, we could start with the number of solutions in max(|x|, |y|, |z|) ∈(2h−1 , 2h ] and then sum over h ≤ log2 H; but even then we would guess that the number of solutions
with max(|x|, |y|, |z|) ≤ H grows only logarithmically. But in fact the rational parametrization bylines through the origin shows that the correct order of growth is H 2/3 . Here the failure of the naı̈veheuristic can be attributed to the singularity of our curve at the origin. In higher dimensions, examples
are known where our heuristic fails for other, subtler reasons.
Still, such failures are not surprising. What is remarkable is how often such a naı̈ve heuristic gives
the correct answer when this answer can be established, and an answer consistent with or close to thepredictions of more refined conjectures and heuristics when the correct answer is not known but the
problem fits into a suitable mathematical framework. In the next few sections we illustrate this by
successively generalizing the problem of solving x2 + y 2 = z 2 until we reach the ABC conjecture.

6.2 Fermat’s “Last Theorem” (FLT): xn + y n = z n
Of the many fruitful generalizations of x2 + y 2 = z 2 , one of the most natural and by far the best
known is the Fermat equation xn + y n = z n for n ≥ 2. Again we seek solutions in nonzero integers,
or equivalently solutions of X n + Y n = 1 in rational numbers X = x/z, Y = y/z. The locus of
X n +Y n = 1 is known as the n-th Fermat curve; Figures 3 and 4 show part of the real locus for n = 3
and the entire real locus for n = 4, and are typical of the visual appearance (albeit not necessarily of
the arithmetic or algebraic geometry) of Fermat curves with n ≥ 3 odd or even respectively.
Y
6

Y
6

- X

- X

Figure 3: X 3 + Y 3 = 1

Figure 4: X 4 + Y 4 = 1

Fermat’s “Last Theorem” (FLT) is the assertion, recorded by Fermat in 1637 and proved by him atleast for n = 4, that for n ≥ 3 there are no solutions of xn + y n = z n in nonzero integers; equivalently,
real field R, and x2 + y 2 = 3z 2 has no nontrivial solution in the field Q3 of 3-adic numbers. Since we live in the real world
rather than the 3-adic world, the former obstruction is more intuitive to us, but both R and Q3 (and more generally Qp for any
prime p) are completions of Q with respect to the corresponding valuations on Q, and decades of experience have shown theadvantage of regarding the real and p-adic valuations of Q on as equal a footing as possible.
At this point we cannot resist another digression. Both x2 + y 2 + z 2 = 0 and x2 + y 2 = 3z 2 are obstructed not just over R
and Q3 respectively, but also over Q2 . It turns out that for any irreducible homogeneous quadratic P (x, y, z) there are at mostfinitely many completions of Q in which there are no nonzero solutions of P (x, y, z) = 0, and that the number — call it ν — ofsuch completions (either real or p-adic) is always even; this is equivalent to Quadratic Reciprocity. Conversely, any finite subsetof {R, Q2 , Q3 , Q5 , Q7 , . . .} of even size can arise this way, a fact that ultimately amounts to the determination of the 2-torsionof the Brauer group of Q. Finally, if ν = 0 then P (x, y, z) = 0 does in fact have nontrivial rational solutions; that is, the Hasse
principle holds for homogeneous quadratics in three variables over Q.

60

that the n-th Fermat curve has no rational points other than (±1, 0) and (0, ±1) (with minus signs
allowed only when n is even). Why should n ≥ 3 behave so differently from n = 2? Let us consult ourheuristic for estimating the expected number of solutions of xn +y n = z n with max(|xn |, |y n |, |z n |) ∈(N/2, N ]. (Every solution (x, y, z) will satisfy this condition with N = 2h for a unique nonnegative
integer h.) As before we write (A, B, C) = (xn , y n , z n ), and observe that A + B = C, and thatthe number of triples (A, B, C) of integers with A + B = C and max(|A|, |B|, |C|) ∈ (N/2, N ]is asymptotically proportional to N 2 . But now we want each of them to be not a square but an n-th
power for some n ≥ 3, and n-th powers get rarer as n increases. Indeed the number of n-th powers in[−N, N ] grows only as N 1/n , so the probability that three integers A, B, C chosen independently anduniformly at random in that range are all n-th powers is asymptotically proportional to N 3((1/n)−1) .
We thus expect roughly N 2+3((1/n)−1) = N (3−n)/n such triples with A + B = C. The exponent(3 − n)/n is positive, zero, or negative according as n < 3, n = 3, or n > 3. Taking N = 2hand summing over h, we thus expect the solutions to be plentiful for n < 3 (the number of solutions
up to N growing as a positive power of N ), sparse for n = 3, and finite in number for n > 3. The
same should be true of primitive3 integral solutions of A0 xn + B0 y n = C0 z n for any fixed choice ofA0 , B0 , C0 , corresponding to rational points on the curve A0 X n + B0 Y n = C0 .
It turns out that each of these predictions is essentially correct. For n = 1 the result is almost trivial.
For n = 2 we saw that, once the curve A0 X 2 + B0 Y 2 = C0 has a rational point P , the lines through P
yield the expected plenty of rational points on the curve. For n ≥ 3 we must appeal to more advanced
and recent results on Diophantine equations. When n = 3, the curve E : A0 X 3 + B0 Y 3 = C0
is a nonsingular cubic plane curve, and thus an elliptic curve assuming it has a rational point P .4
Here it is not so easy to get new rational points, because a typical line through P meets E at two
more points, which in general are not rational. To obtain a new rational point we must use the line
joining two rational points on E, or tangent to one rational point. This is shown in Figure 5 forthe curve with (A0 , B0 , C0 ) = (1, 1, 91): the line through the rational points5 (3, 4) and (6, −5)
meets E again at (9/2, −1/2), and the tangent at (6, −5) meets E again at (−204/341, 1535/341).
Y6

Y = −5 + 36
25 (6 − X)
204 1535
, 341
− 341

(3, 4)



9
1
2, −2

 - X

(6, −5)Figure 5: some rational points on X 3 + Y 3 = 91
3 An integer solution (x, y, z) of a homogeneous polynomial equation p(x, y, z) = 0 is said to be primitive if gcd(x, y, z) =1. Every integer solution other than (0, 0, 0) can be written uniquely as (kx, ky, kz) for some primitive solution (x, y, z) andsome positive integer k.
4 It is known that in characteristic zero such a curve is always isomorphic to one of the more familiar form Y 2 = P (X) for3
some polynomial P3 with distinct roots. See [Si1, Chapter III, §3] for such isomorphisms, and [Si1, Chapter III, §1] for standard
formulas for elliptic curves.
5 The value C = 91 was chosen so that our curve has two simple rational points (3, 4) and (6, −5). This required a simple
0
but nontrivial solution of X 3 + Y 3 = X 0 3 + Y 0 3 . It would have been nice to use the famous “Ramanujan taxicab” example
C0 = 1729 = 13 + 123 = 93 + 103 ; but this would make it hard to draw a clear and accurate Figure 5, because (1, 12) is
too close to an inflection point of E and (10, 9) too close to the middle of the curve. Our example with (3, 4) and (6, −5) relies
instead on another famous
identity 33 + 43 + 53 = 63 , which is tantalizingly reminiscent of 32 + 42 = 52 but alas does notPnngeneralize further: n+2
m=3 m 6= (n + 3) once n > 3.

61

By drawing more lines and tangents we can generate infinitely many rational points on X 3 + Y 3 = 91,
and it can be shown that every rational point can be obtained this way. As one might guess from thecase of (−204/341, 1535/341), the resulting primitive solutions of x3 + y 3 = 91z 3 grow rapidly, andit turns out that the number of primitive solutions with all variables in [−N, N ] is asymptotic only to
R log N for some R > 0. There are similar results for any elliptic curve E. By a famous theorem of
Mordell [Mo] there is a finite list of rational points on E from which all other points can be recovered byrepeatedly drawing chords and tangents through points already known or constructed. More precisely,
Mordell uses the chords-and-tangents construction to give the set E(Q) of rational points on E thestructure of an abelian group,6 and proves that this group is finitely generated. It then follows fromthe Néron-Tate theory of canonical heights that the number of rational points (x/z, y/z) with each ofx, y, z in [−N, N ] is asymptotic to R(log N )ρ/2 , where ρ is the rank of the abelian group E(Q) and R
is a positive constant depending on E. The curve has finitely many rational points if and only if ρ = 0.
It is known that this happens for the cubic Fermat curve X 3 + Y 3 = 1, whose only rational points are
the obvious (1, 0), (0, 1), and the point at infinity (X : Y : 1) = (1 : −1 : 0).
Finally, for n > 3 the curve A0 X n + B0 Y n = C0 is a smooth plane curve of degree at least 4.
Mordell conjectured that (as our heuristics suggest) every such curve has only finitely many rational
points.
At any rate there is no longer a general method for constructing new points out of known ones; eventhe line through two known points, or tangent to one known point, meets the curve in n − 2 more points(allowing points with complex coordinates), and those points need not be rational once n − 2 > 1. For
example, the line X + Y = 1 through the rational points (X, Y ) = (0, 1) and (1, 0) on the Fermat
4
4quartic X
in a pair of Galois-conjugate points, each defined only
√ + Y = 1 meets the√curve again √over Q( −7), namely 12 (1 ± −7), 12 (1 ∓ −7) . More generally, Mordell conjectured that any
algebraic curve of genus at least 2 has only finitely many rational points. (The genus of a curve isa measure of its complexity7 ; an irreducible plane curve of degree d has genus (d − 1)(d − 2)/2 atmost, with equality if and only if the curve is smooth; an elliptic curve has genus 1, and rationallyparametrized curves have genus 0.) Mordell’s conjecture was finally proved by Faltings, who gave two
entirely different proofs [F1, F2]. Like Mordell’s proof of the finite generation of E(Q) for an elliptic
curve E, both of Faltings’ proofs are “ineffective”: Mordell’s proof yields an upper bound on the rank,
and either of Faltings’ proofs yields an upper bound on the number of rational points, but in general
there may be no way to find a list of points and prove that it accounts for all the rational points on
the curve. While much more is known now than at the time of Mordell’s or even Faltings’ proof, thegeneral problems of making those theorems effective remain open.
A final note on Mordell’s and Faltings’ theorems: while they share the mystery of ineffectivity, theproofs are of quite a different flavor. Mordell’s proof for elliptic curves can be traced back to Fermat’s
proof of the case n = 4 of FLT (showing in effect that the elliptic curves Y 2 = X 4 ± 1 associatedto the Diophantine equations x4 ± y 4 = z 2 have rank zero), and can be regarded as the culmination of
Fermat’s work in this direction. On the other hand, Faltings’ proofs, together with the proof of FLT
by Wiles and Taylor [Wil, TW], depend heavily on some of the most abstract and difficult results andtechniques of late twentieth-century number theory; it would take an expository paper at least as longas this one to even give a sense of these methods to a reader not already acquainted with them.

6.3 The Darmon-Granville theorem: xp + y q = z r
Another natural way to generalize the Fermat equation is to allow different exponents, changing xn +y n = z n to xp + y q = z r . Here p, q, r are fixed positive integers that are not necessarily equal, andx, y, z are integer unknowns. Solving this equation is equivalent to solving A + B = C under the
6 While the chord-and-tangent method has been known at least since the time of Fermat, the construction of an abelian grouplaw from it is not obvious. See [Si1, Chapter III, §2] for the details.
7 At this point it is almost obligatory for an expository paper to cite the fact that an algebraic curve of genus g is one whosegraph over C is an orientable surface with g holes; if nothing else, that is one indication that g measures the curve’s complexity.

62

condition that A be a p-th power, B be a q-th power, and C be a r-th power. The Fermat equation withexponent n is the special case p = q = r = n. Applying our heuristic to general (x, y, z), we find that
if A, B, C are random integers with max(|A|, |B|, |C|) ∈ (N/2, N ] then they are respectively p-th,q-th, and r-th powers with probability asymptotically proportional to N ((1/p)−1)+((1/q)−1)+((1/r)−1) ,and thus that of the roughly N 2 solutions of A + B = C in that range we might expect about
N ((1/p)−1)+((1/q)−1)+((1/r)−1) N 2 = N ((1/p)+(1/q)+(1/r)−1)
to yield solutions of xp + y q = z r . As before, the same analysis applies (to the extent we believe it) tothe equation
A0 xp + B0 y q = C0 z r(6.4)
for fixed nonzero A0 , B0 , C0 . This leads us to introduceδ = δ(p, q, r) := 1 −

1 1 1
− − .
p q
r

(6.5)

Our expected number of solutions with max(|A|, |B|, |C|) ∈ (N/2, N ] is now roughly N −δ , and as
before we vary N and expect the solutions to be plentiful, sparse, or bounded according as δ < 0,
δ = 0, or δ > 0. The corresponding values of (p, q, r) are as follows.
Exercise 6.3.1. We have δ(p, q, r) < 0 if and only if one of the following conditions holds: thesmallest of p, q, r equals 1; the two smallest of p, q, r both equal 2; or (p, q, r) is a permutation of(2, 3, 3), (2, 3, 4), or (2, 3, 5). In this case, if min(p, q, r) = 2 then 1/δ is a negative integer. We
have δ(p, q, r) = 0 if and only if (p, q, r) is a permutation of (3, 3, 3), (2, 4, 4), or (2, 3, 6). Otherwise
δ(p, q, r) ≥ 1/42, with equality if and only if (p, q, r) is a permutation of (2, 3, 7).
The new borderline cases (2, 4, 4) and (2, 3, 6) again yield elliptic curves, with equations Y 2 =X 4 ± 1 and Y 2 = X 3 ± 1 in the simplest case A0 = B0 = C0 = 1. It so happens that again eachof these elliptic curves has rank zero, and thus only finitely many rational points. For Y 2 = X 4 ± 1
the only rational points not at infinity are obvious ones with XY = 0; this is equivalent to Fermat’s
result that there are no solutions of x4 ± y 4 = z 2 in nonzero integers. For Y 2 = X 3 ± 1 there is one
additional solution8 32 = 23 + 1, giving rise to a single set of equivalent solutions of x2 + y 3 = z 6
in nonzero integers, namely (x, y) = (3z 3 , −2z 2 ) for nonzero z ∈ Z. For general A0 , B0 , C0 there
may be infinitely many such equivalence classes, but again their minimal representatives will be quite
sparse, with the number of representatives in the range max(|A|, |B|, |C|) ≤ N growing only as amultiple of (log(N )ρ/2 ) (where as before ρ is the rank of the corresponding elliptic curve).
But for general p, q, r our prediction can be very wide of the mark: there are cases where δ > 0 but
solutions are plentiful. For example, the equation x3 + y 4 = z 5 has the solution(x, y, z) = (209952, 11664, 1944) = (25 38 , 24 36 , 23 35 ),(6.6)

with (A, B, C) proportional to (1, 2, 3) — and indeed every integer solution of A + B = C is proportional to (x3 , y 4 , z 5 ) for some (and thus for infinitely many) integer triples (x, y, z). More generally
we have:
Exercise 6.3.2. Suppose the natural numbers p, q, r are pairwise relatively prime, and A0 , B0 , C0 are
any nonzero integers. Then every integer solution of A+B = C is proportional to (A0 xp , B0 y q , C0 z r )for some (and thus for infinitely many) integer triples (x, y, z), and given the initial A, B, C (not allzero) the number of such (x, y, z) with max(|A0 xp |, |B0 y q |, |C0 z r |) ≤ N is asymptotically proportional to N 1/(pqr) as N → ∞. Moreover there are triples (p, q, r) of relatively prime numbers forwhich δ is arbitrarily close to 1.
8 The elliptic curve Y 2 = X 3 + 1 still has rank zero, but with six rational points: one at infinity, one with X = −1, and twoeach with X = 0 and X = 2. The reader can check that no further points are obtained by intersecting the curve with the tangent
line at any of these points, or the line through any two of them. For instance, (X, Y ) = (2, 3) is the third point of intersectionof Y 2 = X 3 + 1 with the line Y = X + 1 through the obvious points (−1, 0) and (0, 1).

63

The exponent 1/(pqr), though usually small, is positive for all p, q, r; hence if p, q, r are pairwise
relatively prime our equation A0 xp + B0 y q = C0 z r has “plentiful solutions” by our standards, evenwhen the value of δ is almost as positive as it can be. This seems to utterly demolish our heuristic,
which suggests that when δ > 0 there should be only finitely many solutions, and moreover that thistendency should be more pronounced the larger δ gets. But even in favorable cases like the “twisted
Fermat curves” A0 xn + B0 y n = C0 z n our heuristic holds only for primitive solutions, those with
x, y, z pairwise relatively prime. Indeed we should not expect the heuristic to hold when x and y havea large common factor, say d, because then A = A0 xn and B = B0 y n are both multiples of dn ,which makes A + B much likelier to be of the form C0 z n than a random number of the same size.
Our construction of plentiful solutions such as (6.6) likewise exploits large common factors. We thus
restrict attention to solutions with (A, B, C) = (A0 xp , B0 y q , C0 z r ) relatively prime.9,10 In this caseour heuristic agrees precisely with the remarkable theorem of Darmon and Granville (1995):
Theorem 1. [DG]: Let p, q, r be natural numbers such that δ(p, q, r) > 0, and let A0 , B0 , C0 be
any nonzero integers. Then there are finitely many triples (x, y, z) of integers with gcd(x, y, z) = 1
satisfying the equation (6.4).
As with FLT and Faltings’ theorem, the proof is alas much too advanced for us to be able to evenoutline the main ingredients here — though we do note that one key step is an application of Faltings’
theorem itself!
Exercise 6.3.3. The Darmon-Granville theorem may seem a bit stronger than what we suggested, because (A, B, C) might still have a common factor coming from the coefficients A0 , B0 , C0 . But given
those coefficients there are only finitely many possible values of d := gcd(A, B, C). Use this to showthat there are also only finitely many equations A1 xp1 + B1 y1q = C1 z1r whose integer solutions satisfying gcd(A1 xp1 , B1 y1q , C1 z1r ) = 1 account for all solutions of (6.4) with gcd(x, y, z) = 1. Therefore
if we knew Darmon-Granville only under the more restrictive hypothesis that A0 xp , B0 y q , C0 z r be
relatively prime, we could deduce the result in the form quoted above.
Seeing that the Darmon-Granville theorem for equation (6.4) generalizes Faltings’ finiteness result
for the case p = q = r of twisted Fermat curves, can we also generalize FLT to the special case A0 =B0 = C0 = 1 of (6.4), finding all solutions of xp + y q = z r with δ(p, q, r) > 0 and gcd(x, y, z) = 1?
Our heuristic analysis suggests that there should be only finitely many such triples (xp , y q , z r ), but wehave no reason to expect that there should be none at all — and we would not be surprised if some ofthem are quite large, especially for those choices of (p, q, r) that make δ positive but small. Note thatthe Darmon-Granville theorem gives finiteness for any particular choice of (p, q, r) but (like Faltings’
theorem vis-a-vis FLT) leaves open the possibility of infinitely many solutions with (p, q, r) varying as
well.
The full answer is still beyond reach, so we report on the known partial results and conjectures.
The simplest example is the identity 1 + 8 = 9 already noted in connection with (p, q, r) = (2, 3, 6);
it yields a solution 1r + 23 = 32 for all r, satisfying δ(2, 3, r) > 0 for all r > 6. Computer searches
reveal 9 more solutions: 132 + 73 = 29 with δ(2, 3, 9) = 1/18; two solutions
25 + 72 = 34 ,35 + 114 = 1222

(6.7)

with {p, q, r} = {2, 4, 5} and δ = 1/20; two solutions
338 + 15490342 = 156133 ,438 + 962223 = 300429072(6.8)

9 We need not specify pairwise relatively prime, because the relation A + B = C forces any factor of two of A, B, C todivide the third.
10 The failure of our naı̈ve heuristic when A, B, C can have large common factors is related to the failure we noted earlier fora singular cubic curve. Here the surface A0 xp + B0 y q = C0 z r is highly singular at the origin, and a solution with A, B, C all
divisible by a high power of p yields a point (x, y, z) on that surface that is close to that singularity in the p-adic metric.

64

with {p, q, r} = {2, 3, 8} and δ = 1/24; and four solutions
27 + 173 = 712 ,

177 + 762713 = 210639282 ,

14143 + 22134592 = 657 ,

92623 + 153122832 = 1137

(6.9)

with {p, q, r} = {2, 3, 7} and the minimal δ value of 1/42. These computations are reported in [DG],with the discovery of the five large solutions credited to Beukers and Zagier. This list is conjectured tobe complete, based both on further computer searches that revealed no other solutions and on various
partial results that prove special cases of the conjecture. In particular it would follow from this conjecture (plus FLT for n = 3) that xp + y q = z r has no solution in integers p, q, r ≥ 3 and relatively
prime integers x, y, z; this is the Tijdeman-Zagier conjecture, for whose solution Andrew Beal offers
a $50,000 prize [Mau].
The most recent of the partial results in the direction of the conjecture that there are no further
solutions with δ(p, q, r) > 0 is [PSS], a tour de force proving that there are no further solutions for{p, q, r} = {2, 3, 7}. This paper also gives an extensive list (Table 1 at the end of the Introduction) oftriples (p, q, r) for which the corresponding result had been proved earlier, including the triples with{p, q, r} = {2, 4, 5} and {2, 3, 8} seen in the other known solutions (6.7, 6.8). Another special caseis Catalan’s conjecture that 8 and 9 are the only consecutive powers of integers, recently proved byMihăilescu [Mi]; this shows that there are no other solutions with x = 1. The proofs of these partial
results call on a vast range of number-theoretical techniques, including classical methods of elementary,
algebraic, and analytic number theory, Galois representations and modularity as in the proof of FLT,
and algebraic geometry of curves. This huge theoretical arsenal is complemented by sophisticatedcomputational and algorithmic tools that are often essential for carrying out algebraic manipulations orfor completing a proof that has been reduced to a finite but nontrivial calculation.
What about δ(p, q, r) < 0, when we expect that the number of relatively prime solutions of (6.4)
with max(|A|, |B|, |C|) ≤ N can grow as a multiple of N −δ as N → ∞? We easily dispose of the casewhere at least one of p, q, r is 1, when we can simply solve (6.4) for the corresponding variable x, y,or z in terms of the other two. So we assume that each of p, q, r is at least 2. In Exercise 6.3.1, we sawthat then −δ = 1/d for some integer d > 0. There are choices of the coefficients A0 , B0 , C0 for which(6.4) has no solutions at all — we already saw the examples x2 + y 2 + z 2 = 0 and x2 + y 2 = 3z 2
with p = q = r = 2. But if we assume that there is at least one solution of (6.4) in relatively prime
integers then Beukers showed [Beu] that the N 1/d guess is correct. Moreover, for each A0 , B0 , C0
there are finitely many polynomial identities in degree 2d that together account for all the relatively
prime solutions, in the same way that the single identity (6.2) accounts for all Pythagorean triples. (Infact the Pythagorean parametrization illustrates the special case A0 = B0 = C0 = 1, p = q = r = 2
of Beukers’ result; note that here δ = −1/2 and both sides of the identity are polynomials of degree 4.)
Unlike the Faltings and Darmon-Granville finiteness results, Beukers’ is effective: at least in principle one can find all the parametrizations by carrying out a computation whose length is boundedby an explicit function of p, q, r, A0 , B0 , C0 . Doing this in practice still requires some work. For thethree exceptional cases where only one of p, q, r equals 2, this work was recently completed by Edwards [Ed]. In particular he gave for the first time the complete solution for {p, q, r} = {2, 3, 5} in thecase A0 = B0 = C0 = 1. There are 27 inequivalent identities, of which the simplest (which Beukers
had already obtained) is X(t)2 + Y (t)3 = Z(t)5 whereX(t) = (t10 + 124 ) (t20 − 122 522 t15 − 124 10006 t10 + 126 522 t5 + 128 ),Y (t) = −t20 − 122 228 t15 − 124 494 t10 + 126 228 t5 − 128 ,Z(t) = 12(−t

11

2

6

(6.10)

4+ 12 11 t + 12 t).

For any m, n ∈ Z we recover an integer solution of x2 + y 3 = z 5 by taking x = n30 X(m/n),
y = n20 Y (m/n), and z = n12 Z(m/n), and these x, y, z are relatively prime if and only if
gcd(m, 6n) = 1

and m 6≡ 2n mod 5.
65(6.11)

For example, m = n = 1 yields 369347901658572 + 2405462393 = 2678285 . To make it suchthat max(|x2 |, |y 3 |, |z 5 |) less than N it is enough to make both |m| and |n| less than some multipleof N 1/60 ; the number of such (m, n) satisfying (6.11) is asymptotically proportional to N 1/30 =N −δ(2,3,5) as expected.
We conclude this section with another scenic detour: a view of two surprisingly pertinent alternativedescriptions of the triples (p, q, r) of integers greater than 1 for which δ(p, q, r) < 0. First, p, q, r satisfy
this condition if and only if there exists a spherical triangle ∆ with angles π/p, π/q, π/r on the unit
sphere Σ, in which case the triangle has area π · (−δ). Second, we have δ(p, q, r) < 0 if and only if thegroup Γ = Γp,q,r with the presentation
Γp,q,r := hα, β, γ | αp = β q = γ r = αβγ = 1i(6.12)

is finite, in which case it has 2d elements, where d = −1/δ as before. The first equivalence follows
from the well-known fact that the sum of the angles of ∆ exceeds π by an amount equal to the areaof ∆. In this case we can take the generators α, β, γ of Γ to be rotations about the vertices of ∆through angles 2π/p, 2π/q, 2π/r, or equivalently the products of pairs of reflections in the edges of ∆.
If we identify Σ with the Riemann sphere CP1 and let t be a complex coordinate on Σ then Γ becomes afinite group of automorphisms of CP1 , which is to say a finite group of fractional linear transformations
t 7→ (at + b)/(a0 t + b0 ). Then for each of our identities X(t)p + Y (t)q = Z(t)r in degree 2d the ratios
X p /Z r , Y q /Z r , etc. are invariant under Γ for a suitable choice of spherical triangle ∆! Moreover, by
Galois theory any such ratio T actually generates the field of Γ-invariant rational functions of t; that
is, C(T ) = (C(t))Γ . For example, when p = q = r = 2 our Pythagorean parametrization (6.2) yields
functions such as (t2 − 1)2 /(2t)2 and (t2 + 1)2 /(2t)2 invariant under the 4-element group isomorphic
with Γ2,2,2 and generated by t ↔ −t and t ↔ 1/t. For (p, q, r) = (2, 3, 5), we have Γ ∼= A5 , the groupof rotational symmetries of a regular icosahedron inscribed in Σ, and the roots of the polynomials11
X, Y, Z of (6.10) are precisely the 30 edge centers, 20 face centers, and 12 vertices of that icosahedron!
When δ(p, q, r) = 0 or δ(p, q, r) > 0 the triangle ∆ is respectively planar or hyperbolic rather
than spherical, and the group Γ = Γp,q,r generated by pairs of reflections in its edges is no longer
finite. But Γ is still intimately connected with xp + y q = z r via automorphisms of Riemann surfaces.
When δ(p, q, r) = 0, we can regard Γ as a group of affine linear transformations t 7→ at + b of C; its
finite-index subgroup of translations (with a = 1) is then a lattice, and the quotient of C by this latticeis the elliptic curve we obtained from xp + y q = z r . When δ = δ(p, q, r) is positive, ∆ is a hyperbolic
triangle of area πδ and Γ is a discrete group of isometries of the hyperbolic plane H; the quotient H/Γ
can be identified with CP1 via a Γ-invariant meromorphic function on H analogous to the functions Tof the previous paragraph, and quotients of H by subgroups of finite index in Γ yield finite extensionsof C(T ) that are used in the proof of the Darmon-Granville theorem and in the solution of some special
cases such as x2 + y 3 = z 7 .

6.4 The ABC conjecture: A + B = C
Masser and Oesterlé noted that a solution of the Fermat equation, or of a natural generalization suchas the equation (6.4) addressed by Darmon and Granville, yields relatively prime numbers A, B, C
(such as xn , y n , z n for a primitive Fermat solution) such that A + B = C and each of A, B, C has
many repeated prime factors. This inspired them to guess a vastly more general constraint on repeatedprime factors in A, B, A + B for coprime integers A, B, and to formulate a precise conjecture on
the nature of this constraint, now known as the ABC conjecture. This conjecture is stated in termsof an arithmetic function called (for reasons whose explanation would take us too far afield here) the“conductor”, defined as follows:
11 Note that X, Y, Z are regarded as homogeneous polynomials of degrees 30, 20, and 12 respectively, so we count t = ∞√among the roots of Z. The other roots of Z are 0 and the ten values of %φ where φ is (1 ± 5)/2 (the golden ratio or its
2
algebraic conjugate) and % is one of the five fifth roots of 12 in C.

66

Definition 2. The conductor N (D) of a nonzero integer D is the product of the (positive) primes
dividing D, counted without multiplicity. Equivalently, N (D) is the largest squarefree factor of N .
Example 6.4.1. N (D1 D2 ) ≤ N (D1 )N (D2 ) for all nonzero integers D1 , D2 , with equality if and only
if they are relatively prime; N (Dn ) = N (D) for all nonzero integers D and n ≥ 1. The following
brief table gives N (D) for 24 ≤ D ≤ 32:
D
N (D)

24
6

25 26
5 26

27 28 29 30 31
3 14 29 30 31

32
2

The size of the integer |D|/N (D) should be regarded a measure of how far D is from being squarefree,that is, of how rich D is in repeated prime factors.
Conjecture 3. (Masser-Oesterlé [Oe]): For every real  > 0 there exists c > 0 such that
N (ABC) > c C 1−(6.13)

holds for all relatively prime natural numbers A, B, C such that A + B = C; equivalently, for every
real  > 0 there exists c > 0 such that
N (ABC) > c max(|A|, |B|, |C|)1−

(6.14)

holds for all relatively prime integers A, B, C such that ±A ± B ± C = 0.
The equivalence is elementary, and the more symmetrical form ±A ± B ± C = 0 will let us avoid
repeating some arguments twice or thrice according to the signs of A, B, C.
In the following exercises, we detail how the ABC conjecture implies “asymptotic FLT” (that is,
FLT for sufficiently large n) as well as its generalizations by Darmon-Granville and Tijdeman-Zagier,and then give an equivalent formulation in terms of the “ABC exponent”, and explain why the  in
(6.13,6.14) cannot be removed.
Exercise 6.4.1. The ABC conjecture applied to (A, B, C) = (A0 xp , B0 y q , C0 z r ) implies the DarmonGranville theorem; moreover, for any p, q, r such that δ = δ(p, q, r) > 0 and any positive  < δ, theinequality (6.13) with an explicit value of c yields an explicit upper bound on relatively prime integers
x, y, z such that A0 xp + B0 y q = C0 z r .
Exercise 6.4.2. The ABC conjecture implies the Tijdeman-Zagier conjecture with at most finitely many
exceptions; moreover, for any positive  < 1/12 the inequality (6.13) with an explicit value of c yields
an explicit upper bound on xp , y q , z r in any counterexample to the conjecture.12
Exercise 6.4.3. The ABC conjecture for any  < 1 implies that Fermat’s Last Theorem holds for all
but finitely many exponents n. Again, an explicit value of c yields an explicit n0 such that FLT holds
for all n ≥ n0 .
Exercise 6.4.4. The ABC conjecture for any  < 1 implies that any finitely generated multiplicative
subgroup G of Q∗ contains only finitely many solutions (s, s0 ) of s+s0 = 1. [Choose generators for G,
and let S be the set of primes that divide the numerator
or denominator of at least one generator; thenQs + s0 = 1 yields A + B = C with N (ABC) | p∈S p.]
Remark. For this problem, as with the first exercise in this list, the finitude of solutions is already atheorem, without assuming ABC or any other unproved conjecture. Better yet, explicit upper bounds
have been given on C as a function of N (ABC) — whereas no such bound is known for the DarmonGranville theorem without an ABC hypothesis. Still, the proved bounds are much worse than what
would follow from (6.13); see below.
12 The bound 1/12 can be raised to 1/6 because Bruin showed [Br] that there are no solutions of x3 +y 3 = z 4 or x3 +y 3 = z 5
in relatively prime integers.

67

Exercise 6.4.5. For relatively prime natural numbers A, B, C such that A + B = C, define the ABC
exponent θ(A, B, C) byθ(A, B, C) := (log C) (log N (ABC))(so that C = N (ABC)θ(A,B,C) ); for example θ(1, 8, 9) = log 9/ log 6 = 1.226+. Set
Θ := lim sup θ(A, B, C),(A,B,C)the limsup running over all triples (A, B, A + B) of natural numbers. Then the ABC conjecture isequivalent to Θ ≤ 1. In fact the ABC conjecture is equivalent to Θ = 1, because it is elementary thatΘ ≥ 1 (for instance we may take (A, B) = (1, 2r − 1) with r → ∞).
Remark. If it is true that lim sup θ(A, B, C) = 1 then the convergencemust be very slow: it is known√
that there are infinitely many examples of θ(A, B, C) > 1 + c/ log C for some universal constant c >0; and it is expected, based on probabilistic heuristics such as applied earlier to A0 xp + B0 y q = C0 z r ,that in fact θ(A, B, C) − 1 > (log C)−ϑ holds infinitely often for all ϑ > 1/3, but only finitely often
for each ϑ < 1/3. In particular, the ABC conjecture is consistent with those heuristics. The largest
numerical value known for θ(A, B, C) is 1.6299+, for 2 + 310 109 = 235 (found by Eric Reyssat in1987). See [Ni] for other large θ(A, B, C).
Exercise 6.4.6. The inequality (6.13) cannot hold for  = 0 and any positive value of c0 . (One way toprove this is to find for each α > 0 a natural number r such that 3α |2r − 1.)
The ABC conjecture, like FLT, is formulated over Z but has an equivalent statement over Q obtainedby considering ratios of the variables. If A + B = C, consider F = A/C, so 1 − F = B/C. Bothfractions are in lowest terms because A, B, C are assumed relatively prime. The conductor N (A) isthe product of the primes p such that F ≡ 0 mod p, and likewise N (B) is the product of the primes p
such that F ≡ 1 mod p. As for N (C), that is the product of primes p for which F mod p cannot befound in Z/pZ because the denominator C vanishes mod p. Since in this case p - A, we say that these
are the primes such that “F ≡ ∞ mod p”. Hence N (ABC), the LHS of the ABC conjecture (6.13),
is the product of primes p such that F mod p is one of 0, 1, ∞. The RHS is c C 1− , in which C issimply the denominator of F . This assumes that A, B, C are positive, that is, that 0 < F < 1; inthe general case we replaced C by max(|A|, |B|, |C|) (see (6.14)), so now we replace the denominatorof F by the height h(F ). By definition, the height of a rational number m/n with gcd(m, n) = 1 ismax(|m|, |n|). This need not exactly equal max(|A|, |B|, |C|), but is within a factor of 2, which canbe accommodated by changing the constant c of (6.14). Thus the ABC conjecture is equivalent to theassertion that for every  > 0 there exists c > 0 such that, for all F ∈ Q, the product of the primes atwhich F reduces to 0, 1, or ∞ is at least c h(F )1− .
Geometrically, the reduction of FLT to ABC in Exercise 6.4.3 amounts to applying the ABC conjecture to the value of the rational function F = (x/z)n = X n on the n-th Fermat curve. This succeeds
because F and 1 − F have multiple poles and zeros (some defined only over an algebraic closure Q)— that is, the preimages of 0, 1, ∞ under F have large multiplicities, which makes the total number ofpreimages counted without multiplicity small compared to the degree of F as a rational function on the
curve. It turns out that here the degree is n2 , and the number of preimages is 3n, which is less than n2
once n > 3, and indeed less than δn2 once n > 3/δ. When we try to generalize this argument to rational points on a general algebraic curve X , we find that it is rare for there to be a rational function F
on X whose degree exceeds the size of F −1 ({0, 1, ∞}) by a large factor, so we cannot usually expect todeduce Mordell’s conjecture (finiteness of rational points) for X from an ABC inequality with  near 1.
But Belyi [Bel] shows how to construct a function F satisfying deg(F ) > # F −1 ({0, 1, ∞}) whenever X is a curve of genus at least 2 defined by an equation with coefficients in Q, and then Mordell’s
conjecture follows from ABC with  sufficiently small [El1]. Recall that Faltings already proved thisconjecture twice without any unproved hypothesis, but the proofs are ineffective; the argument in [El1]68shows that the ABC conjecture with effective constants c would yield a completely effective finitenessresult for rational points on X .
Many other consequences of the ABC conjecture are known, ranging from elementary special cases(there are only finitely many integers N such as N = 4, 5, 7 for which N ! + 1 is a perfect square) toapplications that give unexpected connections with other problems in number theory. A striking example is Silverman’s application to Wieferich primes, that is, primes p for which 2p−1 ≡ 1 mod p2 ,such as 1093 and 3511. (Note that the congruence always holds mod p by Fermat’s little theorem. In
1909 Wieferich proved [Wie] that a FLT counterexample xp + y p = z p with p - xyz for some prime pwould imply 2p−1 ≡ 1 mod p2 .) Such primes are expected to be very rare; indeed none is known otherthan 1093 and 3511, and any further such prime must exceed 1.25 · 1015 according to computations reported by Richard McIntosh (http://www.loria.fr/∼zimmerma/records/Wieferich.
status). But it is not even known that the set of non-Wieferich primes is infinite! Silverman [Si2]proves the infinitude of non-Wieferich primes under the hypothesis of the ABC conjecture, and shows
further that this conjecture implies that for every integer α 6= 0, ±1 there exist constants cα , xα suchthat for all x > xα there are at least cα log xα primes p < x satisfying αp−1 6≡ 1 mod p2 .
Unfortunately a proof of the ABC conjecture still seems a very distant prospect; it is even much
too hard to prove the existence of any  < 1 for which the inequality (6.13) holds for some c > 0. To
show just how far we are, consider the situation suggested by Exercise 6.4.4: we know N = N (ABC),and want all possible (A, B, C). Let S be the set of primes dividing N . Then the inequality (6.13) forany  < 1 gives an upper bound on solutions of A + B = C in relatively prime integers all of whose
prime factors are contained in S. (This is often called the “S-unit equation”, because it is equivalent
to solving a + b = 1 in rational numbers (a, b) = (A/C, B/C) that are units in the ring Z[1/N ]obtained from Z by inverting all the primes in S.) In particular, there should be only a finite numberof solutions. This result is known [La1], but already far from trivial. It was not much harder to givean explicit bound on the number of solutions [LM], and by now there are bounds depending only on
the size of S, as in [Ev]. But that still gives no control over the size of the largest solution, which iswhat the ABC conjecture addresses. Stewart and Tijdeman gave such a bound in [ST], and the bound
was recently improved by Stewart and Yu [SY]. But even the best bounds remain exponential: thelogarithm of C is only known to be bounded by a multiple of N 1/3 (log(N ))3 . Even these results canbe useful; for instance the Stewart-Tijdeman bound log C = O(N 15 ) is already enough to compute inpractice the full solution of the S-unit equation when S is not too large (see for instance [dW]). But
the known results are still very weak compared with the inequalities that the ABC conjecture predicts
and that we need for applications such as the Tijdeman-Zagier conjecture and explicit upper bounds inthe Darmon-Granville theorem.

6.5 Mason’s theorem: A(t) + B(t) = C(t)A curious feature of the ABC conjecture is that not only does it seem very hard to prove but it is not at
all obvious how one might try to disprove the conjecture. If FLT were false, a single counterexample
would expose the falsity; likewise for the Catalan and Tijdeman-Zagier conjectures, or the Riemannhypothesis and its variants. But there can be no single counterexample for the ABC conjecture, evenfor a specific value of , because the inequality (6.13) can accommodate any given triple (A, B, C)
by simply decreasing c . Likewise for the formulation of the conjecture in terms of ABC exponentsθ(A, B, C): a single example may break the record for the maximal θ, but has no bearing on Θ which isdefined as a lim sup of θ(A, B, C). Proving that the conjecture is false would require the existence of an
infinite family of (A, B, C)’s whose ABC exponents approach a limit greater than 1 (or approach ∞),just as we had to construct an infinite family such as {(1, 2r − 1, 2r )}∞r=1 to prove Θ ≥ 1.
The most natural families to try arise from identities A(t) + B(t) = C(t) relating polynomials
A, B, C ∈ Z[t], not all constant. Recall that we already used such polynomials to construct infinitely
many Pythagorean triples, or relatively prime solutions of x2 + y 3 = z 5 ; in effect we solved these69

Diophantine equations in Z[t], then specialized to t ∈ Q and multiplied by powers of the denominatorof t to recover integer solutions. Similarly, from polynomials A(t), B(t), C(t) satisfying A + B = C forwhich D := max(deg(A), deg(B), deg(C)) is positive we get a family of integer solutions A, B, C as
follows: for any pair (m, n) of relatively prime integers we take(A, B, C) = nD (A(m/n), B(m/n), C(m/n)).

(6.15)

Thus A, B, C are homogeneous polynomials of degree D in (m, n). If A, B, C have repeated factors
then so do A, B, C, and with enough repeated factors we can hope to get a sequence withlim sup θ(A, B, C) > 1.
We must assume that A(t), B(t), C(t) are relatively prime as polynomials, else A, B, C will havea common factor for most choices of (m, n). This also means that D is the degree of the quotient
F = A/C ∈ Q(t) as a rational function of t. Conversely, if the polynomials have no common factors
then gcd(A, B) is bounded above,13 so dividing each of our triples (A, B, C) of (6.15) by its greatest
common divisor yields relatively prime solutions of A + B = C with asymptotically the same ABC
exponent as the ratio
log max(|A|, |B|, |C|)log max(|A|, |B|, |C|)=
(6.16)
log N (ABC)
log(N (A)N (B)N (C))
that we would compute if A, B, C were relatively prime.
The numerator of this ratio is easy to estimate: it is D log h(m, n) + e, whereD = max(deg(A), deg(B), deg(C))
as above, h(m, n) is the height |max(m, n)| of (m, n) (or of the rational number m/n as before), ande is an error of bounded absolute value. What of the denominator? Let us try some examples using
polynomial identities that we have already encountered. If
(A, B, C) = (t2 − 1)2 , (2t)2 , (t2 + 1)2
as in (6.2), then D = 4 and we get (A, B, C) = ((m2 − n2 )2 , (2mn)2 , (m2 + n2 )2 ) (the squares ofthe entries of the Pythagorean triple (6.2)), and then N (ABC) is a factor of (m2 − n2 )2mn(m2 + n2 ).
Hence N (ABC) is bounded above by a multiple of h(m, n)6 . We can save two factors of h(m, n)
in various ways, for instance by making (m, n) = (1, 2r ) as in Exercise 6.4.5; but that still leaves
both numerator and denominator of (6.16) asymptotic to 4 log h(m, n), giving the same lower boundof 1 on Θ that we obtained in that Exercise. Might we do better with the more complicated example(A, B, C) = (X(t)2 , Y (t)3 , Z(t)5 ), where X, Y, Z are the polynomials of (6.10)? Now D = 60 andA, B, C are respectively a square, a cube, and a fifth power, so N (ABC) is bounded by a multiple ofh(m, n)30+20+12 = h(m, n)62 . Again we can save a factor h(m, n)2 thanks to the factor mn of C, butthat still brings our bound on N (ABC) only down to a multiple of h(m, n)60 = h(m, n)D , and again
we fail to improve on Θ ≥ 1.
Q
In general, suppose A factors as A0 i xei i where A0 is a scalar and theQxi are distinct irreducible
polynomials.
Let xi = ndeg xi xi (m/n). Then A = nD A(m/n) = A0 ne∞ i xei i , where e∞ := D −PDi ei deg(xi ) is the multiplicity of n as a factor of the homogeneous polynomial n A(m/n) (which
may also be regarded as the “order of vanishing at t = ∞” of A when
Q A is regarded
Q as a polynomialof degree D). Hence N (A) is bounded by a constant multiple of i xi or n i |xi | according as
e∞ = 0 or e∞ > 0. Each |xi | is in turn bounded by a constant multiple of (h(m, n))deg xi , and ofP
course |n| ≤ h(m, n). It follows that N (A) ≤ h(m, n)νD (A) where νD (A) = νD,∞ (A) + i deg xi
13 By the Euclidean algorithm for polynomials there exist X, Y ∈ Z[t] such that AX − BY = d for some nonzero d ∈ Z,
and then gcd(A, B) | nD d for all m, n ∈ Z. Repeating this argument with A, B replaced by the relatively prime polynomialstD A(1/t), tD B(1/t) yields a nonzero integer d0 such that gcd(A, B) | mD d0 . Thus if gcd(m, n) = 1 then gcd(A, B) | dd0 .

70and νD,∞ (A) = 0 or 1 according as e∞ = 0 or e∞ > 0. More succinctly, νD (A) is the numberof solutions of F(t) = 0 in CP1 , counted without multiplicity (note in particular that e∞ > 0 if andonly if F(∞) = 0). We define νD (B) and νD (C) likewise, and observe that they are the numbersof solutions in CP1 of F(t) = 1 and F(t) = ∞. Putting these together we find that N (A, B, C)
is bounded by a constant multiple of h(m, n)ν where ν = νD (A) + νD (B) + νD (C) is the size of
F−1 ({0, 1, ∞}). Moreover, if at least two points in F−1 ({0, 1, ∞}) are rational then we can save an
extra factor of h(m, n)2 as we did before; in fact we expect to save this factor in any case, because
there are about H 2 choices of (m, n) with h(m, n) ∈ (H/2, H], and it is not too hard to show that infact this h(m, n)2 saving is available for all nonconstant rational functions F. In other words, we canmake the denominator of (6.16) no larger than (ν − 2) log h(m, n) + e0 , where e0 is another boundederror.
Combining our estimates and letting h(m, n) → ∞, we find that the polynomial identity A+B = C
will yield a disproof the ABC conjecture if ν < D + 2. We have already given several examples ofν = D + 2, and there are many others, some of which are very easy to construct (try (A, B, C) =(1, tD − 1, tD ) for instance). Might we attain ν < D + 2 if we are just a little more clever, or lookharder? This is where Mason’s theorem enters:
Theorem 4. [Mas]: If F ∈ C(t) is a rational function of degree D > 0 on CP1 then F−1 ({0, 1, ∞})has cardinality at least D + 2.
This ruins our hope for an easy refutation of the ABC conjecture. Viewed more positively, it isevidence for the truth of the conjecture, and indeed can be viewed as an “ABC theorem” for polynomials
or rational functions. To make the comparison explicit, we again take logarithms in the conjectured
inequality (6.13) to write it as log N (ABC) > (1 − ) log C − log(1/c ). We saw that for polynomials
ν and D play the roles of log N (ABC) and log C respectively. Thus Mason’s theorem is an evenstronger statement, because the troublesome terms − log C and − log(1/c ) in the lower bound forlog N (ABC) have been replaced by the helpful +2 in the lower bound on ν.
Moreover, while the ABC conjecture seems intractable at present, Mason’s theorem can be proved
easily. There are several related routes, all exploiting the idea of detecting multiple roots of a polynomial or rational function using its derivative — a tool not available for integers or rational numbers.
The route we choose uses the logarithmic derivative, for which it will be convenient to assume that ∞is not a preimage of 0, 1, or ∞. We ensure this by applying to t a fractional linear transformation thatmoves all the preimages of {0, 1, ∞} away from infinity.
Proof. Fix a number t0 not in F−1 ({0, 1, ∞}), and let F1 (t) = F(t0 + (1/t)), a rational function alsoof degree D and with the same number of preimages of {0, 1, ∞} as F, none of which are at infinity.
Let ν0 , ν1 , ν∞ be the number of preimages of 0, 1, ∞ respectively. Let Λ be the logarithmic derivative
F01 /F1 . Then Λ is not identically zero because F1 is nonconstant, and Λ has a simple pole (that is, hasa denominator with a simple root) at each preimage of 0 or ∞, regardless of its multiplicity. Hencethe denominator of Λ has degree ν0 + ν∞ . Any root of F01 − 1 of multiplicity e is a root of Λ ofmultiplicity e − 1. Summing over the roots, we find the the numerator of Λ has at least D − ν1 roots
counted with multiplicity, and therefore has degree at least D − ν1 . But the difference between the
denominator’s and numerator’s degrees
P∞ is the order of vanishing of Λ at infinity, which is at least 2 (tosee this, expand F1 at infinity as i=0 ai t−i = a0 + a1 t−1 + a2 t−2 + a3 t−3 + · · · with a0 6= 0, andcalculate F01 = −a1 t−2 − 2a2 t−3 − 3a3 t−4 − · · · ). Hence D − ν1 ≤ ν0 + ν∞ − 2, which is equivalent
to the desired inequality ν0 + ν1 + ν∞ ≥ D + 2.
Since the numerator of the derivative or the logarithmic derivative of A/C is (up to sign) the Wronskian
A C= AC0 − A0 C,
W2 (A, C) = det
A0 C0
the proof can also be formulated in terms of Wronskians. The key fact that F and F − 1 have the same
derivative then corresponds to the identity W2 (A, C) = W2 (A − C, C), which holds because W2 (·, ·) is
71

bilinear and alternating, and forces W2 (A, C) to vanish at multiple zeros of B. Also equivalent, though
not as transparently so, is the proof obtained by applying the Riemann-Hurwitz formula to F. Thisapproach explains the “+2” in Mason’s inequality as the Euler characteristic of CP1 , and generalizes
to rational functions F of degree D
 > 0 on other compact Riemann surfaces, for which Mason finds
the inequality # F−1 ({0, 1, ∞}) ≥ D + χ = D + 2 − 2g, where g is the genus and χ the Euler
characteristic of the surface. This is why the rational functions F constructed by Belyi cannot satisfy
deg(F ) > # F −1 ({0, 1, ∞}) unless g ≥ 2. For an elliptic curve we have g = 1, so deg(F ) =#F −1 ({0, 1, ∞}) is possible, and if the elliptic curve has positive rank then its rational points yield
another kind of infinite family of (A, B, C) triples with lim sup θ(A, B, C) ≥ 1 (such as (x3 , y 3 , 91z 3 )for primitive solutions of x3 + y 3 = 91z 3 ); but the points are too sparse for us to prove that the limsup
strictly exceeds 1, and again we come just short of a disproof of the ABC conjecture.

6.6

A Putnam problem: minding our P ’s and Q’s

The last problem of the 1956 William Lowell Putnam Mathematical Competition asks [GGK, p.47]:
The polynomials P (z) and Q(z) with complex coefficients have the same set of numbers
for their zeros but possibly different multiplicities. The same is true of the polynomials
P (z) + 1 and Q(z) + 1. Prove that P (z) ≡ Q(z).
As noted in [GGK, p.431], it must be assumed that at least one of P and Q is not constant, else theclaim is false. We thus assume max(deg(P ), deg(Q)) > 0, and by symmetry may take m = deg(P ) ≥deg(Q) = n. The claim is clearly true if P has distinct roots, because then Q = cP for some c ∈ C,
and if λ is any root of P + 1 then 0 = Q(λ) + 1 = cP (λ) + 1 = −c + 1 implies c = 1. Likewise
if P + 1 has distinct roots. We must then contend with the case that P and P + 1 both have multiple
roots — and we know already that the derivative P 0 = (P + 1)0 detects multiple roots of either P or
P + 1. We proceed as in [GGK, p.431–432]. Let λ1 , . . . , λr be the distinct roots of P (and thus alsoof Q), and µ1 , . . . , µs the distinct roots of P + 1 (and thus also of Q + 1). By an argument we can now
recognize as the special case of Mason’s theorem in which F is a polynomial — and which would failif m = 0 were allowed — we have m − 1 = deg(P 0 ) ≥ 2m − r − s, whence r + s ≥ m + 1. But eachroot of P or P + 1 is also a root of P − Q, a polynomial of degree at most m. Therefore P − Q is thezero polynomial, and we are done.
The corresponding statement for integers instead of polynomials would be that a positive integer n
is determined uniquely by the sets (without the multiplicities) of prime factors of n and of n + 1, thatis, by the conductors N (n) and N (n + 1). We might expect that this should be false, because the proofin the polynomial case hinges on an inequality stronger than can be true for integers. Indeed there areinfinitely many counterexamples, the smallest with natural numbers being n = 2 and n0 = 8 (this is yet
another appearance of 1 + 8 = 9), which begins the infinite family {n, n0 } = {2m − 2, 2m (2m − 2)}(m = 2, 3, 4, . . .). Still, such examples seem quite rare; an exhaustive search finds that the only casewith 0 < n, n0 < 108 not of the form {2m − 2, 2m (2m − 2)} is {75, 1215} (with N (75) = N (1215) =15 and N (76) = N (1216) = 38). When we allow also negative integers, the identity N (−n) = N (n)gives an involution {n, n0 } ↔ {−1 − n, −1 − n0 } on the set of solutions. Modulo this involution, we
find one more infinite family {2m + 1, −(2m + 1)2 } (m = 1, 2, 3, . . .), and one more sporadic pair in(−108 , 108 ), namely {35, −4375}. The infinite families intersect at {2, −4, 8} and {−3, 3, −9}, which
may be the only three-element subsets of Z mapped to a single point under n 7→ (N (n), N (n + 1)).
Might we generalize the Putnam problem to rational functions F ? Since a polynomial is just a
rational function with F −1 ({∞}) = {∞}, we might guess that more generally if F and G are nonconstant rational functions with complex coefficients that, when considered as maps from the Riemannsphere CP1 to itself, satisfy F −1 ({w}) = G−1 ({w}) for each of w = 0, 1, ∞, then F = G. (In the
Putnam problem, F and G would be the polynomials P + 1 and Q + 1.) Alas this natural guess is false.
72An explicit counterexample is14
F (z) =(z − 1)3 (z + 3),
16z

G(z) = h(−3/z) =(z − 1)(z + 3)3,
16z 3

with F (z) − 1 = (z − 3)(z + 1)3 /16z and G(z) − 1 = (z − 3)3 (z + 1)/16z 3 . Here F and G are
rational functions of degree 4. Is this the smallest possible? It is probably much harder to completely
describe all counterexamples, or even to decide whether there any with deg(F ) 6= deg(G).

6.7 Further problems and results
In number theory most things that can be done in Q or Z generalize, with some additional effort, tonumber fields K (finite-degree field extensions of Q) and their rings OK of algebraic integers. This istrue of the ABC conjecture, which can be naturally formulated over any K or OK , and has much thesame consequences there as we saw over Q or Z. Much of the extra effort in making this generalizationarises because OK need not have unique factorization, so some solutions in K of A + B = C may not
be proportional to any solution in relatively prime elements of OK . Thus it is more natural to formulate
the conjecture in terms of the ratio F = A/C, which is invariant under scaling (A, B, C). Briefly, we
replace N (ABC) in the LHS of (6.13) or (6.14) by the product of the norms of all prime ideals of OK
at which F is congruent to one of 0, 1, ∞, and in the RHS we take the (1 − )-th power of the heightof F , appropriately defined, rather than of C or of max(|A|, |B|, |C|). See [Vo, p.84] for the details.
Mason’s theorem still defeats attempts at easy disproofs — recall that the coefficients of the rational
function F were allowed to be arbitrary complex numbers.
More subtle is the question of how the constant c in the ABC conjecture should depend on K. In
the context of Mason’s theorem, if we replace C(t) by a finite-degree extension we get the function
field of a compact Riemann surface of some genus g, and then the lower bound D + 2 on the sizeof F−1 ({0, 1, ∞}) is lowered by 2g. Granville and Stark [GS] propose an anologous “uniform ABC
1/[K:Q]
conjecture”, in which the LHS of (6.13) or (6.14) is multiplied by |disc(K/Q)|
and then theconstant c in the RHS is independent of K. Remarkably, they then show that this uniform ABC
conjecture
implies the long-standing conjecture that the class number of an imaginary quadratic field√Q( −d) (with d > 0 a squarefree integer) is bounded below by a constant multiple of d1/2 / log d, andthus that the Dirichlet L-function attached to an odd character has no “Siegel-Landau zero” (a zero swith 1 − s  1/ log(d); the nonexistence of such zeros is an important special case of the RiemannHypothesis for such L-functions). The proof uses special values of modular functions
arising from√
elliptic curves with complex multiplication by the ring of algebraic integers in Q( −d).
Finally we consider the generalization to more than three variables, to integers satisfying ±A ±B ± C ± D = 0 and beyond. In each case we ask: Given max(|A|, |B|, |C|, . . .), how small can
the product N (A)N (B)N (C) · · · get? As before we must assume that the integers have no common
factor. With more than three variables, it no longer follows that they are relatively prime in pairs,
but we must at least assume that no proper sub-sum of ±A ± B ± C ± · · · vanishes, to avoid suchtrivialities as 2r + 1 − 2r − 1 = 0. It is then known that an upper bound on N (A)N (B)N (C) · · ·implies an upper bound on max(|A|, |B|, |C|, . . .), but again this known bound is much too large forour purpose. Even in the special case A = A0 wn , B = B0 xn , etc. we have a difficult question: How
14 The reader who got this far may well wonder where this counterexample comes from. It arises naturally in the theory ofQ
2πinτ )24 ,elliptic modular functions. For τ in the upper half-plane
H, let η(τ ) be the Dedekind eta function eπi/12 ∞n=1 (1−eQ2n )/(1 + q 2n−1 ) where η = η(kτ ) and q = eπiτ . Then λand define λ(τ ) = 16(η22 η1/2 /η13 )8 = 16q ∞
(1
+
q
n=1
k
generates the field of modular functions invariant under the ideal hyperbolic triangle group Γ(2), and takes the values 0, 1, ∞ at
the cusps of that group. The function F expresses λ in terms of the generator −3(η3 /η1)10 (η1/2 η2 /η3/2 η6 )4 of the modular
functions for Γ(2) ∩ Γ0 (3), and thus gives explicitly the map from the corresponding modular curve to the modular curve X(2)corresponding to Γ0 (2). The coordinate λ of X(2) parametrizes elliptic curves E : Y 2 = X(X − 1)(X − λ) with all their
2-torsion points rational; z parametrizes 3-isogenies E → E 0 between pairs of such curves; and the involution z ↔ −3/z takes
the isogeny E → E 0 to the dual isogeny E 0 → E. See [El2].

73

are the nontrivial primitive solutions of A0 wn + B0 xn + C0 y n = D0 z n distributed? Our heuristics
suggest that solutions should be plentiful for n < 4 (if there is a nonzero solution to begin with), sparsefor n = 4, and bounded for n > 4. Likewise for N variables, with critical exponent n = N .
Unfortunately this guess is at best close to the truth. Euler already found a polynomial solution for
w4 +x4 = y 4 +z 4 , giving plentiful solutions for that equation, starting with 1334 +1344 = 594 +1584 .
There is even a polynomial family of solutions of w5 + x5 = y 5 + z 5 , though sadly not over Q:
w, x = 2t ± (t2 − 2),y, z = 2t ± i(t2 + 2).

(6.17)For n = 6 one can still obtain infinitely many primitive solutions for some choices of (A0 , B0 , C0 , D0 ),using the polynomial identity(t2 + t − 1)3 + (t2 − t − 1)3 = 2t6 − 2.
Indeed let (A0 , B0 , C0 , D0 ) = (α3 , β 3 , 2, 2). Then if there are infinitely many rational solutions(t, u, v) of
t2 + t − 1 = αu2 , t2 − t − 1 = βv 2(6.18)
then each yields a rational solution (u, v, 1, t) of A0 w6 + B0 x6 + C0 y 6 = D0 z 6 , and thus a primitive
integer solution by clearing common factors. Now it can be shown that (6.18) is an elliptic curve,
which has positive rank if it has a single rational point with t ∈/ {0, ±1, ∞}. The simplest such (α, β)is (5, 1) with t = 2, giving 125 + 1 + 2 = 2 · 26 . The next few t values for (α, β) = (5, 1) are −82/19,−148402/91339, and −10458011042/1213480199, giving the solutions15(31, 19, 89, 82),(5009, 91339, 165031, 148402),(4363642319, 1213480199, 10981259039, 10458011042).
Note that, unlike the ABC conjecture, our naı̈ve guess for A0 wn + B0 xn + C0 y n = D0 z n wasdisproved by polynomial identities. Thus even Mason’s theorem has no good analogue here. One canuse a 3×3 Wronskian to get an “ABCD theorem”, and likewise for more variables, but these inequalities
are no longer sharp. For example, if (w, x, y, z) is a nontrivial solution in C[t] of wn + xn = y n + z nthen one can show that n < 8 by counting roots of W3 (wn , xn , y n ), but it is not known whether n = 6
or n = 7 can occur, nor whether all nontrivial solutions for n = 5 are equivalent with (6.17).
Can we salvage from our predicament a conjecture that is both plausible and sharp? Lang [La2]suggested that such conjectures should still be true “on a nonempty Zariski-open set”, that is, when we
exclude variables that satisfy some algebraic condition. This may well be true, though the possibilityof an unpredictable exceptional set makes Lang’s conjectures even harder to test. As an indication of
the power of these conjectures, we conclude by citing one striking application. Recall that Mordell
conjectured, and Faltings proved, that an algebraic curve of genus g > 1 over Q has only finitely many
rational points. The conjecture and proofs are silent on how the number of points can vary with thecurve. But Caporaso, Harris, and Mazur showed [CHM] that Lang’s conjectures imply a uniform upperbound B(g), depending only on g, on the number of rational points of any genus-g curve over Q!

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