David KRUMM
Squarefree parts of polynomial values
Tome 28, no 3 (2016), p. 699-724.
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Journal de Théorie des Nombresde Bordeaux 28 (2016), 699–724

Squarefree parts of polynomial values
par David KRUMM

Résumé. Étant donné un polynôme non constant séparable f (x)à coefficients entiers, nous considérons l’ensemble S constitué desparties sans facteurs carrés de toutes les valeurs rationnelles def (x), et étudions son comportement modulo un nombre premier.
Ayant fixé un nombre premier p, nous déterminons des conditions
nécessaires et suffisantes pour que S contienne un élément divisible
par p. Nous conjecturons que si p est suffisamment grand, alors S
contient une infinité de représentants de chaque classe résiduelle
non nulle modulo p. Nous prouvons cette conjecture quand f (x)est de degré 1 ou 2. Si f (x) est de degré 3, ou s’il est de degré
4 avec une racine rationnelle, la preuve de la conjecture utilise laconjecture de parité pour les courbes elliptiques. Pour les polynômes de degré arbitraire, un analogue local de la conjecture est
prouvé en utilisant des résultats standard de la théorie des corps declasse. Des résultats numériques sont aussi inclus qui confirment
la version globale de la conjecture.
Abstract. Given a separable nonconstant polynomial f (x) withinteger coefficients, we consider the set S consisting of the squarefree parts of all the rational values of f (x), and study its behavior
modulo primes. Fixing a prime p, we determine necessary andsufficient conditions for S to contain an element divisible by p.
We conjecture that if p is large enough, then S contains infinitely
many representatives from every nonzero residue class modulo p.
The conjecture is proved by elementary means assuming f (x) has
degree 1 or 2. If f (x) has degree 3, or if it has degree 4 and has
a rational root, the conjecture is shown to follow from the parity
conjecture for elliptic curves. For polynomials of arbitrary degree,
a local analogue of the conjecture is proved using standard resultsfrom class field theory, and empirical evidence is given to support
the global version of the conjecture.

Manuscrit reçu le 18 juillet 2014, révisé le 16 juillet 2015, accepté le 16 juillet 2015.
Mathematics Subject Classification. 11A07, 11R09, 11G05, 11G30.
Mots-clefs. Squarefree part, hyperelliptic curve, quadratic twist, congruent number.

700David Krumm

1. Introduction
The squarefree part of a nonzero rational number r, which we denote hereby S(r), is the unique squarefree integer d such that r/d is a square in Q.
To any separable polynomial f (x) ∈ Z[x] of positive degree we associate
the following set of squarefree integers:
S(f ) = {S(f (r)) : r ∈ Q and f (r) 6= 0}.
An alternate definition of the set S(f ) can be made in more geometricterms. Let C be the hyperelliptic curve defined by the equation y 2 = f (x),and for any squarefree integer d, let Cd denote the quadratic twist of C byd; i.e., the hyperelliptic curve defined by dy 2 = f (x). It may be the casethat the curve Cd has rational points at infinity or affine rational points(x, y) with f (x) = 0; we will refer to these as trivial rational points. With
this terminology,S(f ) = {squarefree d : Cd has a nontrivial rational point}.
For any prime number p we consider the problem of determining which
residue classes modulo p are represented in the set S(f ). A simple criterion
for deciding whether S(f ) contains a multiple of p can be readily obtained— see Theorem 2.1. The case of nonzero residue classes is not nearly as
simple; in fact, most of this article is devoted to studying that case. Our
main goal is to provide evidence in support of the following:
Conjecture 1.1. For all but finitely many primes p, the set S(f ) contains
infinitely many elements from every nonzero residue class modulo p.
The initial motivation for this work was an observation made in the process of studying quadratic points on the modular curves Y1 (N ). Consider,for example, the quadratic number fields K such that the curve Y1 (18) has
a K-rational point1 . As shown in [16, Thm. 2.6.5], if we let
f (x) = x6 + 2x5 + 5x4 + 10x3 + 10x2 + 4x + 1,√
then every such field has the form K = Q( d) for some d ∈ S(f ). Moreover,every number d ∈ S(f ) satisfies d ≡ 1 (mod 8) and d ≡ 0 or 1 (mod 3).
In view of this result it becomes natural to ask, for any integer n, which
residue classes modulo n are represented in the set S(f ). This question canof course be asked not only for this particular polynomial f (x), but for any
polynomial. In this article we restrict attention to prime values of n andto separable polynomials in order to simplify various arguments. Thus, wearrive at the problem of determining which residue classes modulo a prime
p are represented in a set of the form S(f ).
1 Questions about quadratic points on this curve arise in connection to elliptic curves overquadratic fields, and also to dynamical properties of quadratic polynomials over quadratic fields;see [7, §3.11].

Squarefree parts of polynomial values

701

Another instance in which this problem becomes of interest is a question concerning congruent numbers. Recall that a positive integer is called
congruent if it is the area of a right triangle with rational sides. The setof congruent numbers is closely related to the set S(f ) for the polynomial
f (x) = x3 − x. Indeed, it is easy to see that a positive integer is congruent if and only if its squarefree part is congruent, and furthermore, it iswell known that a positive squarefree integer d is congruent if and only
if d ∈ S(f ); see [15, Chap. 1]. The Birch and Swinnerton-Dyer conjecture
would imply that S(f ) contains every positive squarefree integer d ≡ 5, 6
or 7 (mod 8). This result has not been proved unconditionally, but partial results have been achieved. Notably, Monsky [21] has shown that S(f )contains infinitely many elements from each of the residue classes 5, 6,
7 modulo 8. Now, for any integer n one might ask which residue classes
modulo n contain a congruent number. With regards to this question, thefollowing is shown in §3.
Theorem 1.1. For every prime number p there exist infinitely many
congruent numbers divisible by p. Furthermore, assuming the Parity Conjecture, every residue class modulo p contains infinitely many congruent
numbers.
For polynomials of small degree we can give the best evidence supporting
our main conjecture. The next result summarizes our work in §3.
Theorem 1.2. Conjecture 1.1 holds unconditionally if f (x) has degree 1
or 2. If f (x) has degree 3, or if it has degree 4 and has a rational root, thenthe conjecture follows from the Parity Conjecture.
For polynomials of higher degree we can only show that there is no local
obstruction to the conjecture. Letting C denote the hyperelliptic curvey 2 = f (x), the conjecture can be rephrased as follows: if p is a large enoughprime, then for every integer m coprime to p there exist infinitely manysquarefree integers d such that d ≡ m (mod p) and Cd has a nontrivial
rational point. Collecting the main results of §4 we obtain the following
local version of the conjecture.
Theorem 1.3. Suppose that f (x) has odd degree, or that it has even degreeand some irreducible divisor of f (x) has abelian Galois group. Then there isan effectively computable constant N such that the following holds for every
prime p ≥ N : if m is any integer coprime to p, then there exist infinitely
many squarefree integers d ≡ m (mod p) such that the curve Cd has anontrivial point over every completion of Q.
Remark. We expect that the Galois group condition imposed when f (x)has even degree is unnecessary, although our arguments do make essential
use of this assumption.

702

David Krumm

As an improvement to Conjecture 1.1 it would be desirable to have a precise quantitative statement about how the elements of S(f ) are distributed
among the residue class modulo p. Given a prime p and an integer m, weconsider the function
D(t) = #{d ∈ S(f ) : |d| ≤ t and d ≡ m (mod p)},and ask what the asymptotic behavior of D(t) is as t → ∞. A similar
question has been studied by other authors, but without restricting d toany particular residue class: Stewart-Top [30, Thm. 2] show that#{d ∈ S(f ) : |d| ≤ t}  t1/(g+1) /(log t)2 ,where g is the genus of the hyperelliptic curve y 2 = f (x), and Granville [10,Conj. 1.3] conjectures that if g ≥ 2, then there is a positive constant cf suchthat#{d ∈ S(f ) : |d| ≤ t} ∼ cf · t1/(g+1) .
Another useful function to consider is defined as follows. Recall that theheight of a rational number r = a/b expressed in lowest terms is the number
H(r) = max{|a|, |b|}. For any real number t ≥ 1, letS(f, t) = {S(f (r)) : f (r) 6= 0 and H(r) ≤ t}.
With p and m as above, we then define
E(t) = #{d ∈ S(f, t) : d ≡ m (mod p)}
and ask how this function behaves as t → ∞. A related question was
answered by Poonen in [24, Thm. 3.5], though again disregarding residue
classes: under the assumption of the abc conjecture, if f (x) has degree ≥ 2,then#{S(f (1)), S(f (2)), . . . , S(f (t))} ∼ t.
If results analogous to those of Stewart-Top or Poonen could be obtained forthe functions D(t) or E(t), we would have a proof (perhaps conditional on
the abc conjecture) and a more precise version of Conjecture 1.1. However,
it is not clear whether the methods of these authors can be modified toyield asymptotics for our counting functions.
This article is organized as follows. In §2 we give necessary and sufficientconditions for the set S(f ) to contain an element divisible by p. A discussionof Conjecture 1.1 for polynomials of degrees 1-4 is given in §3. In §4 weprove a local analogue of the conjecture for polynomials of arbitrary degree.
Finally, §5 contains empirical evidence supporting Conjecture 1.1.
Remark. Some of the material in this article appears in the author’s Ph.D.
thesis [16].

Squarefree parts of polynomial values703

2. The residue class of 0
Let f (x) ∈ Z[x] be a nonconstant separable polynomial. We show inthis section how to decide, for most primes p, whether the residue classof 0 modulo p is represented in the set S(f ). This particular residue classis easier than others to deal with because there is a simple criterion todecide whether p divides the squarefree part of a nonzero rational number r.
Indeed, we have p|S(r) if and only if ordp (r) is odd. Here, ordp denotes thestandard p-adic valuation.
It will be convenient from this point on to distinguish between two typesof primes. Throughout this article we say that an odd prime p is goodfor f (x) if the reduced polynomial f¯(x) ∈ Fp [x] has the same degree asf (x) and has nonzero discriminant. This condition on p implies that thehyperelliptic curve y 2 = f (x) has good reduction modulo p; see [19, p. 464,
Ex. 1.26].
Lemma 2.1. Let R(f ) be the set of all primes q that are good for f (x) andsuch that f (x) has a root modulo q. Then R(f ) is an infinite set.
Proof. The Chebotarev Density Theorem implies that there are infinitely
many primes q such that f (x) has a root modulo q. (See [3, Thm. 2], forinstance.) The result now follows by noting that all but finitely many primes
are good for f (x).
Lemma 2.2. For any finite subset T of R(f ) there exists d ∈ S(f ) suchthat d is divisible by every prime in T .
Proof. Let t be the product of all the primes in T , and let
A = {n ∈ Z : f (n) 6= 0, t|f (n), and gcd(t, f 0 (n)) = 1}.
We claim that A is nonempty. By definition of R(f ), for every prime q ∈ T
there is an integer nq such that q|f (nq ). Moreover, since q is good for f (x),every root of f (x) modulo q must be a simple root; hence, q does not divide
f 0 (nq ). Let n be an integer satisfying n ≡ nq (mod q) for all q ∈ T , andf (n) 6= 0. (The existence of n is guaranteed by the Chinese Remainder
Theorem.) For every q ∈ T we havef (n) ≡ f (nq ) ≡ 0 (mod q) and f 0 (n) ≡ f 0 (nq ) 6≡ 0 (mod q).
Therefore n ∈ A, proving that A is nonempty. We partition A into subsets
B and C defined by
B = {n ∈ A : ordq (f (n)) ≡ 1 (mod 2) for all q ∈ T }and C = A \ B. Note that if n ∈ B, then the number d = S(f (n)) is
divisible by every prime in T . Hence, the proof will be complete if we showthat B is nonempty. Since A is nonempty, at least one of B and C must

David Krumm704

be nonempty. Assuming C is nonempty, we will show that B must also benonempty, and this will conclude the proof.
Given n ∈ C, let W ⊆ T consist of all primes q ∈ T such that ordq (f (n))
is even. For every q ∈ W , write ordq (f (n)) = 2sq with sq ≥ 1. For every
prime q ∈ T \ W , set ordq (f (n)) = rq , which is an odd positive integer.
Define v ∈ Z by the formulav=

Y

q 2sq −1 ·q∈WYq rq +1 .

q∈T \W

Using a Taylor expansion we see thatf (n + v) = f (n) + f 0 (n) · v + z · v 2
for some integer z. Now, for every q ∈ W we haveordq (f (n)) = 2sq and ordq (z · v 2 ) ≥ 2ordq (v) = 4sq − 2 ≥ 2sq .
Moreover, by definition of A, q does not divide f 0 (n), and so
ordq (f 0 (n) · v) = ordq (v) = 2sq − 1 < 2sq .
Therefore, ordq (f (n + v)) = 2sq − 1 is odd. By a similar argument we seethat for primes q ∈ T \ W , ordq (f (n + v)) = rq is also odd. Thus, f (n + v)
has odd and positive valuation at every prime in T . Finally, for every q ∈ T ,f 0 (n + v) is congruent to f 0 (n) modulo q, and is therefore not divisible byq. We conclude that n + v ∈ B, showing that B is nonempty.

We record a consequence of the proof of Lemma 2.2 for use in a later
section.
Lemma 2.3. Let q be a good prime for f (x) such that f (x) has a root
modulo q. Then there exists an integer n such that ordq (f (n)) is odd and q
does not divide f 0 (n).
Proof. Following the proof of Lemma 2.2 with T = {q}, this result is thestatement that the set B is nonempty.
Proposition 2.1. With notation as in Lemma 2.2, there exist infinitely
many d ∈ S(f ) such that d is divisible by every prime in T .
Proof. Let t be the product of all the primes in T , and fix any primeq ∈ R(f ) \ T . Applying Lemma 2.2 to the finite subset T ∪ {q} of R(f )we see that there is an element dq ∈ S(f ) that is divisible by qt. Since dq is
divisible by q, the map q 7→ dq is necessarily finite-to-one (as every elementof S(f ) is only divisible by finitely many primes). By Lemma 2.1, the set
R(f ) \ T is infinite, so its image in S(f ) must also be infinite. Hence, there
are infinitely many elements of the form dq in S(f ). By construction, every
number dq is divisible by t, and hence by every prime in T .
Squarefree parts of polynomial values

705

Theorem 2.1. Let p be a good prime for f (x).
(1) If f (x) has odd degree, then there are infinitely many d ∈ S(f ) suchthat p|d.
(2) Suppose f (x) has even degree.
(a) If there exists d ∈ S(f ) such that p|d, then f (x) has a root
modulo p.
(b) Conversely, if f (x) has a root modulo p, then there are infinitely many d ∈ S(f ) such that p|d.
2g+1
Proof. Suppose first that f (x) has odd degree, and write f (x) = i=0
ai xi .
If g = 0, the result follows by noting that for any prime q 6= p the equationf (r) = pq has a rational solution, so that pq ∈ S(f ). As q varies over all
primes different from p, we obtain infinitely many elements of S(f ) thatare divisible by p. Assume now that g ≥ 1. The argument in this case isbased on evaluating f (x) at values 1/pn for odd n. LetP

F (x) = x2g+2 · f (1/x) = x(a2g+1 + a2g x + · · · + a0 x2g+1 ).
Since p is good for f (x), p does not divide a2g+1 . Hence, for every positive
integer n,
ordp (F (pn )) = n.
In particular, if n is odd, then ordp (F (pn )) is odd, so p divides the squarefreepart of F (pn ). Letting d = S(F (pn )), we then havep|d and d = S(f (1/pn )) ∈ S(f ).
Thus, we have shown that the set D = {S(F (pn )) : n is odd} is containedin S(f ), and every number in D is divisible by p. We claim that D is
infinite, which will prove part (1) of the theorem. Note that if d ∈ D, say
d = S(F (pn )), then the Diophantine equation dy 2 = F (x) has an integral
solution (with x = pn ). Since g ≥ 1, Siegel’s theorem [11, Thm. D.9.1]
implies that this equation has only a finite number of integral solutions(x, y). Hence, d can only be the squarefree part of F (pn ) for finitely many
values of n. It follows that the set D is infinite, proving our claim.
To prove part (2), suppose that f (x) has even degree and write f (x) =P2g+2i
i=0 ai x . Let F (x, y) ∈ Z[x, y] be defined by
F (x, y) = y 2g+2 · f (x/y) = a2g+2 x2g+2 + a2g+1 x2g+1 y + · · · + a0 y 2g+2 .
If S(f ) contains an element divisible by p, then there is a rational number r
such that f (r) 6= 0 and the integer d = S(f (r)) is divisible by p. Writingr = a/b with a and b coprime integers, we haved = S(f (a/b)) = S(F (a, b)).
Hence, there is an integer s such that F (a, b) = ds2 . We claim that p
cannot divide b. For suppose that p|b, and reduce the equation ds2 = F (a, b)

David Krumm

706

modulo p; we obtaina2g+2 · a2g+2 ≡ F (a, b) ≡ ds2 ≡ 0 (mod p).
However, this is impossible because p does not divide a2g+2 (as p is goodfor f (x)), and p does not divide a (since p|b and a is coprime to b). This
proves the claim. The equationb2g+2 · f (a/b) = ds2
therefore takes place in the local ring Z(p) , so we may reduce the equation
modulo p to conclude that f¯(a/b) = 0, and hence f (x) has a root modulo
p. This proves part 2(a) of the theorem.
Conversely, suppose that f (x) has a root modulo p. Then p ∈ R(f ), so
we may apply Proposition 2.1 to the set T = {p} ⊂ R(f ), and thus obtainthat there are infinitely many d ∈ S(f ) such that p|d. This proves 2(b). Remark. In [31, Lem. 2], Top shows that if f (x) ∈ Z[x] is a nonconstantseparable polynomial, then there are infinitely many primes p such that theresidue class of 0 modulo p is represented in the set S(f ). In the case thatf (x) has odd degree, Theorem 2.1 provides a stronger statement. Indeed,
the theorem implies that for all but finitely many primes p, the class 0 mod p
is represented in S(f ), and in fact is represented by infinitely many elementsof S(f ).
Having given a simple criterion for deciding whether the residue class of
0 modulo p is represented in S(f ), we will henceforth restrict attention tononzero classes.
3. The case of degrees 1-4
For every positive integer n we define a statement A(n) as follows.
Statement A(n). Let f (x) ∈ Z[x] be a separable polynomial of degree n,
and let p be a good prime for f (x). Then for every integer m coprime to p
there exist infinitely many elements d ∈ S(f ) such that d ≡ m (mod p).
We show in this section that A(1) and A(2) hold unconditionally, andthat A(3) is implied by the Parity Conjecture for elliptic curves over Q.
Furthermore, still assuming the Parity Conjecture, we show that A(4) holds
for polynomials having a rational root. In order to obtain these results it
will be convenient to work with somewhat stronger statements I(n) andP(n).
Statement I(n). Let h(x) ∈ Z[x] be a separable primitive2 polynomialof degree n, and let p be a good prime for h(x). Then for every integer
m coprime to p there exist infinitely many primes q ∈ S(h) such thatq ≡ m (mod p).
2 A polynomial is called primitive if the g.c.d. of its coefficients is equal to 1.

Squarefree parts of polynomial values

707

Lemma 3.1. For every positive integer n, I(n) implies A(n).
Proof. Assume that I(n) holds. Let f (x) be a separable polynomial of degreen, let p be a good prime for f (x), and let m be any integer coprime top. We must show that there are infinitely many elements d ∈ S(f ) suchthat d ≡ m (mod p). By factoring out the greatest common divisor of thecoefficients of f (x), we may write f (x) = δs2 · h(x) with δ squarefree andh(x) primitive of degree n. Note that since p is good for f (x), it is also goodfor h(x). Applying Statement I(n) to h(x) we see that there exist infinitely
many primes q ∈ S(h) such that q ≡ δ −1 m (mod p). Here, δ −1 denotes themultiplicative inverse of δ modulo p. Fix any such prime q that does notdivide δ. The integer d = δq is then squarefree and congruent to m modulo
p. Since q ∈ S(h), there are rational numbers r and t such that h(r) = qt2
and h(r) 6= 0. Thenf (r) = δs2 · h(r) = δq(st)2 = d(st)2 ,so d = S(f (r)) ∈ S(f ). Since we have infinitely many choices for q,
this construction yields infinitely many numbers d ∈ S(f ) such that d ≡m (mod p).
Statement P(n). Let h(x) ∈ Z[x] be a separable polynomial of degree n
with square leading coefficient, and let p be a good prime for h(x). Then forevery integer m coprime to p there exist infinitely many primes q ∈ S(h)such that q ≡ m (mod p).
Lemma 3.2. For every positive integer n, P(n) implies A(n).
Proof. Assume that P(n) holds. Let f (x) be a separable polynomial of
degree n, let p be a good prime for f (x), and let m be any integer coprime
to p. We must show that there are infinitely many elements d ∈ S(f ) suchthat d ≡ m (mod p). Let δ be the squarefree part of the leading coefficientof f (x), and define h(x) = δ · f (x). Note that p is good for h(x), andthat the leading coefficient of h(x) is a square. Applying statement P(n)to h(x) we see that there exist infinitely many primes q ∈ S(h) such thatq ≡ δ −1 m (mod p). (Note that δ has a multiplicative inverse modulo p
because p does not divide the leading coefficient of f (x).) Fix any suchprime q that does not divide δ. The integer d = δq is then squarefree andcongruent to m modulo p. Since q ∈ S(h), we can write qs2 = h(r) for somerational numbers r, s with h(r) 6= 0. Thend(s/δ)2 = δq(s/δ)2 = h(r)/δ = f (r),so d = S(f (r)) ∈ S(f ). Since we have infinitely many choices for q,
this construction yields infinitely many numbers d ∈ S(f ) such that d ≡m (mod p).
708

David Krumm

Lemma 3.3. Let a1 , . . . , an be pairwise coprime integers, and let x1 , . . . , xn
be integers with gcd(ai , xi ) = 1 for all i. Then there exist infinitely many
prime numbers q such that q ≡ xi (mod ai ) for all i.
Proof. By the Chinese Remainder Theorem, there is an integer N suchthat N ≡ xi (mod ai ) for all i. Note that N is coprime to each ai , andthus coprime to the number a = a1 · · · an . Dirichlet’s theorem on primes inarithmetic progressions [14, p. 251, Thm. 1] then implies that there exist
infinitely many primes q that are congruent to N modulo a. Clearly, every
such prime q satisfies q ≡ xi (mod ai ) for all i.
Proposition 3.1. Statement A(1) holds.
Proof. By Lemma 3.1 it suffices to show that I(1) holds. Let f (x) be aprimitive polynomial of degree 1, so that we can write f (x) = ax + b with
gcd(a, b) = 1. Let p be a good prime for f (x), and let m be any integer
not divisible by p. Since p does not divide a, Lemma 3.3 implies that there
exist infinitely many primes q satisfying q ≡ m (mod p) and q ≡ b (mod a).
By construction, every such prime has the form q = b + na = f (n) for somen ∈ Z, and so q = S(f (n)) ∈ S(f ). This shows that there are infinitely
many primes q ∈ S(f ) such that q ≡ m (mod p), and I(1) is proved.
In order to prove that A(2) holds we will need the following classicalresult; see [14, p. 273] for further details on this theorem.
Theorem of Legendre. Let a, b, c be nonzero integers that are squarefree,pairwise coprime, and not all positive nor all negative. Then the equation
ax2 + by 2 + cz 2 = 0
has a nontrivial integral solution if and only if the following conditions are
satisfied:• −bc is a square modulo a;• −ac is a square modulo b;• −ab is a square modulo c.
Proposition 3.2. Statement A(2) holds.
Proof. By Lemma 3.2 it suffices to show that P(2) holds. Let f (x) be aquadratic polynomial with square leading coefficient, so that f (x) has theform f (x) = a2 x2 + bx + c for some integers a, b, c. We assume that thediscriminant ∆ = b2 −4a2 c is nonzero, and write ∆ = δs2 with δ a squarefree
integer and s ∈ Z. Letting p be a good prime for f (x) and m an integer
coprime to p, we must show that there exist infinitely many primes q ∈ S(f )such that q ≡ m (mod p).
Let F (x, y) ∈ Z[x, y] be the binary quadratic form defined by
F (x, y) = y 2 · f (x/y) = a2 x2 + bxy + cy 2 .

Squarefree parts of polynomial values

709

A simple calculation shows that for every prime q we haveq ∈ S(f ) ⇐⇒ F (x, y) represents q over Q.
Thus, it suffices to show that there are infinitely many primes q ≡m (mod p) that are rationally represented by F (x, y). The form F (x, y)
is equivalent (over Q) to the diagonal form G(x, y) = x2 − δy 2 . Indeed,letting
 

X
Y

=

a
0

b
2a
s
2a

! 

x
y

we have G(X, Y ) = F (x, y). Since equivalent forms represent the samevalues, the proof will be complete if we show that there are infinitely many
primes q ≡ m (mod p) that are rationally represented by G(x, y). Since p
is a good prime for f (x), p is coprime 8δ. Hence, by Lemma 3.3 there exist
infinitely many primes q satisfyingq ≡ m (mod p) and q ≡ 1 (mod 8δ).
Letting q be any such prime, we claim that q is represented by G(x, y) over
Q. One can verify using Quadratic Reciprocity that δ is a square modulo
q; Legendre’s theorem then implies that the equationx2 − δy 2 − qz 2 = 0
has a nontrivial integral solution, say (x0 , y0 , z0 ). If δ 6= 1, then we musthave z0 6= 0 since δ is squarefree. In this case we can divide by z02 to obtain
G(x0 /z0 , y0 /z0 ) = q. If δ = 1, then it is trivial to see that G(x, y) represents
q: for instance,
q+1 q−1
,
= q.
G
2
2
This proves our claim and hence the proposition.



We turn now to consider the statement A(3). For this statement we do
not have an unconditional proof as was the case for A(1) and A(2); however,
we can bring the machinery of elliptic curves to bear on the problem, andthus provide compelling evidence that A(3) should hold. Recall that theParity Conjecture for elliptic curves states that the analytic and algebraic
ranks of an elliptic curve over Q must have the same parity. (See the survey
in [26, Chap. 4] for more details.) We will need to use a different versionof this conjecture which relates the rank of an elliptic curve to the rankof a quadratic twist of the curve. For a statement of this conjecture in theliterature, see [9, p. 4].

David Krumm

710

Parity Conjecture for quadratic twists. Let E be an elliptic curve overQ with conductor NE , and let d be a squarefree integer coprime to 2 · NE .
Then(−1)rank Ed (Q) = (−1)rank E(Q) · χd (−NE ),√where χd is the quadratic Dirichlet character associated to the field Q( d).
√Recall that if D is the discriminant of the field Q( d), then the characterχd : (Z/DZ)× → {±1}can be defined using the Kronecker symbol:
 Dχd (n) =.
n(See [22, p. 296] for a definition of this symbol.) In particular, χd has thefollowing properties:• χd (−1) = sign(d)2• If d ≡ 1 (mod 4), then χd (2) = (−1)(d −1)/8 .
 • For any odd prime q - d, χd (q) is the Legendre symbol dq .
Lemma 3.4. Let E/Q be an elliptic curve, p an odd prime not dividing the
conductor of E, and m an integer coprime to p. If the Parity Conjectureholds, then there exist infinitely many squarefree integers d ≡ m (mod p)
such that the twist Ed has positive rank.
Proof. Let
NE = 2e ·

v
Y
pei
i

i=1

be the factorization of the conductor of E, and set ε = (−1)1+rank E(Q) . By
Lemma 3.3, there exist infinitely many primes q satisfyingq ≡ ε (mod 8p1 · · · pv ) and q ≡ εm (mod p).
Fix any such prime q. Letting d = ε · q, we haved ≡ 1 (mod 8p1 · · · pv ) and d ≡ m (mod p).
Note that d is squarefree and coprime to 2 · NE . The properties of thecharacter χd imply thatχd (−NE ) = χd (−1)χd (2)ev  ei
Y
d
i=1

pi

= χd (−1) = sign(d) = ε.

Hence, by the Parity Conjecture,(−1)rank Ed (Q) = (−1)rank E(Q) · ε = (−1)rank E(Q) (−1)1+rank E(Q) = −1.
It follows that Ed has odd, hence positive, rank. Since we have infinitely
many choices for q, this argument yields infinitely many squarefree integersd ≡ m (mod p) such that Ed has positive rank.
Squarefree parts of polynomial values

711

Lemma 3.5. Let C/Q be a hyperelliptic curve of genus 1, p an odd primeof good reduction for C, and m an integer coprime to p. If the Parity
Conjecture holds, then there exist infinitely many squarefree integers d ≡m (mod p) such that the Jacobian of Cd has positive rank.
Proof. Let E = Jac(C), which is an elliptic curve over Q. Since C has good
reduction modulo p, then E also has good reduction modulo p. Hence, pdoes not divide the conductor of E; see [28, p. 256]. By Lemma 3.4, there
exist infinitely many squarefree integers d ≡ m (mod p) such that Ed has
positive rank. The result now follows by noting that Ed = Jac(Cd ). (Thiscan be deduced from the construction of the Jacobian of a genus 1 curve;see [1] or [5, Chap. 20].)

Proposition 3.3. Statement A(3) follows from the Parity Conjecture.
Proof. Let f (x) ∈ Z[x] be separable of degree 3, let p be a good prime
for f (x), and let m be any integer coprime to p. We must show, assuming
the Parity Conjecture, that there are infinitely many elements d ∈ S(f )such that d ≡ m (mod p). Let E be the elliptic curve over Q defined by
the equation y 2 = f (x). Since p is good for f (x), p is an odd prime ofgood reduction for E. Hence, by Lemma 3.5, the Parity Conjecture implies
that there exist infinitely many squarefree integers d ≡ m (mod p) suchthat the Jacobian of Ed has positive rank. However, for every such d, Ed
is an elliptic curve and thus isomorphic to its Jacobian; therefore, Ed has
positive rank. In particular, Ed must have a nontrivial rational point, andso d ∈ S(f ).
Remark. In the proof of Proposition 3.3 we only need the existence of a
nontrivial rational point on the twist Ed , while the Parity Conjecture (viaLemma 3.4) produces a seemingly much stronger result, namely that Ed
has positive rank. It is natural to wonder whether one can avoid recourse to
the conjecture by only proving the existence of a nontrivial torsion point on
Ed . Unfortunately, this approach will not work. It follows from a theoremof Silverman [27, Thm. 6] that if E is any elliptic curve over Q, then all
but finitely many twists Ed have the property that the torsion subgroup of
Ed (Q) contains only 2-torsion points. Since 2-torsion points on an elliptic
curve are – using our terminology – trivial rational points, this leaves only afinite number of twists that might have a nontrivial rational torsion point.
Hence, we cannot expect by this method to produce the infinitely many
integers d required to prove A(3).
We now discuss our main conjecture in the case of polynomials of degree
4. Though we expect that the statement A(4) holds, it does not seem immediate to deduce it from the Parity Conjecture. In particular, if the proofof Proposition 3.3 is followed with a polynomial f (x) of degree 4 instead of

712David Krumm

degree 3, the result will be a statement weaker than A(4): instead of obtaining twists Cd with a nontrivial rational point, we would obtain twists
Cd whose Jacobian has a nontrivial rational point. There is, nevertheless,one instance in which we can deduce A(4) from the Parity Conjecture:
Proposition 3.4. Assume the Parity Conjecture. Then Statement A(4)holds if we restrict attention to polynomials f (x) having a rational root.
Proof. Let f (x) ∈ Z[x] be a separable polynomial of degree 4 with a rationalroot, let p be a good prime for f (x), and let m be any integer coprime to p.
We must show that there are infinitely many elements d ∈ S(f ) such thatd ≡ m (mod p). Let C be the hyperelliptic curve of genus 1 defined by theequation y 2 = f (x). Note that every quadratic twist of C has a rational
point with y = 0. By Lemma 3.5 there exist infinitely many squarefree
integers d ≡ m (mod p) such that the Jacobian of Cd has positive rank.
Since Cd has a rational point, it can be given the structure of an elliptic
curve, and is therefore isomorphic to its Jacobian. Hence, Cd has infinitely
many rational points. In particular, it has a nontrivial rational point, sod ∈ S(f ).
We conclude this section with an application to congruent numbers.
Theorem 3.1. For every prime number p there exist infinitely many congruent numbers divisible by p. Furthermore, assuming the Parity Conjecture, every residue class modulo p contains infinitely many congruent numbers.
Proof. If p = 2, this follows (unconditionally) from Monsky’s result citedin the introduction. Now fix an odd prime p. Recall that a positive squarefree integer d is congruent if and only if d ∈ S(f ), where f (x) = x3 − x.
The discriminant of f (x) is 4, so p is good for f (x). Since f (x) has odd
degree, Theorem 2.1 implies that there are infinitely many elements of S(f )that are divisible by p. Note that f (−x) = −f (x), so that the set S(f ) is
closed under taking additive inverses. Hence, we can conclude that there
are infinitely many positive numbers in S(f ) divisible by p.
To complete the proof we must show, assuming the Parity Conjecture,that every nonzero residue class [m] modulo p contains infinitely many
congruent numbers. The rank of the elliptic curve E defined by the equationy 2 = x3 − x is 0, so the proof of Proposition 3.3 can be used to produce
infinitely many negative numbers d ∈ S(f ) such that d ≡ −m (mod p).
For any such d we have (−d) ∈ S(f ), −d ≡ m (mod p), and −d > 0. Thisshows that there are infinitely many congruent numbers in [m].
Squarefree parts of polynomial values

713

4. A local analogue
Let f (x) ∈ Z[x] be a separable nonconstant polynomial. In the previous
section we provided evidence for Conjecture 1.1 assuming that f (x) has
degree ≤ 4. For polynomials of higher degree we cannot prove our conjecture
or even show that it would be implied by standard conjectures in arithmetic
or Diophantine geometry. Hence, we consider instead a weaker statement:
our main goal in this section is to show that there is no local obstruction
to Conjecture 1.1.
Let C be the hyperelliptic curve y 2 = f (x), let p be a prime, and let mbe an integer coprime to p. Our conjecture states that if p is large enough,then there exist infinitely many squarefree integers d ≡ m (mod p) suchthat the curve Cd has a nontrivial rational point. We will prove a related
statement in which the global condition that Cd have a nontrivial rational
point is replaced by the local condition that it have a nontrivial point overevery completion of Q.
For the convenience of the reader, we recall here a few standard results
to be used.
Hasse-Weil bound. Let X be a smooth, irreducible, projective curve of
genus g over a finite field F` . Then√|#X(F` ) − (` + 1)| ≤ 2g `.
Proof. See [12, §9.2].

Hensel’s Lemma. Let ` be a prime number, and let P (t1 , . . . , tn ) ∈Z` [t1 , . . . , tn ]. Suppose α ∈ Fn` is such that P (α) = 0 and ∇P (α) 6= 0.
Then there exists x ∈ Zn` such that P (x) = 0 and x = α.
Proof. See [25, p. 15, Cor. 1].

Lemma 4.1 (Squares in `-adic fields). Let x ∈ Q∗` and write x = `n u,
where n ∈ Z and u ∈ Z×
` . Then x is a square in Q` if and only if n is evenand the following holds:• If ` is odd, then u is a square in F` .
• If ` = 2, then u ≡ 1 (mod 8).
Proof. See p. 17, Thm. 3 and p. 18, Thm. 4 in [25].

Lemma 4.2. Let g be the genus of C. There exists an integer N0 such thatevery prime ` ≥ N0 satisfies(a) ` is good for f (x); and(b) for every smooth projective curve X/F` of genus g, #X(F` ) ≥ 2g + 5.

714David Krumm

Proof. For (a) to hold we take N0 to be larger than every prime dividing
the discriminant or the leading coefficient of f (x). For (b), the existence of
N0 is guaranteed by the Hasse-Weil bound; for instance, it suffices to haveN0 ≥ 4g 2 + 6g + 4.
Lemma 4.3. With N0 as in Lemma 4.2, let d be a squarefree integer andlet ` ≥ N0 be prime.
(1) If f (x) has odd degree, then the curve Cd has a nontrivial point
defined over Q` .
(2) Suppose f (x) has even degree. If ` - d, or if `|d and f (x) has a root
modulo `, then Cd has a nontrivial point defined over Q` .
Proof. Suppose first that ` does not divide d. Since ` is good for f (x), theequation dy 2 = f¯(x) defines a hyperelliptic curve of genus g over F` , which
we denote by Ced . By the definition of N0 we have #Ced (F` ) ≥ 2g + 5. The
curve Ced can have at most 2 points at infinity, so it must have at least
2g + 3 affine points defined over F` . At most 2g + 2 of these points canbe of the form (α, 0), so there must be a point (α, β) with β 6= 0. Let
F (x, y) = dy 2 − f (x) ∈ Z` [x, y] and note thatF (α, β) = 0 and ∇F (α, β) = −f¯0 (α), 2dβ 6= (0, 0)
since f¯(x) has no repeated root. Hensel’s Lemma implies that the point(α, β) lifts to a point (a, b) ∈ Z2` with db2 = f (a). Moreover, since β 6= 0,
we must have b 6= 0, so (a, b) is a nontrivial point in Cd (Q` ). Thus we haveshown that, regardless of what the degree of f (x) is, if ` does not divide d,then Cd has a nontrivial point defined over Q` .
Now suppose that ` divides d.
Case 1 : f (x) has odd degree. Let F (x) = x2g+2 · f (1/x) ∈ Z[x], and let
a be the leading coefficient of f (x). A simple calculation shows that we canwrite F (ad) = db, where b ∈ Z satisfies b ≡ a2 (mod `). Thus, b is a nonzero
square modulo `. By Lemma 4.1 there exists a (nonzero) element y ∈ Q`
such that y 2 = b. We have F (ad) = dy 2 , and sof (1/ad) = F (ad)/(ad)2g+2 = dy 2 /(ad)2g+2 = d(y/(ad)g+1 )2 .
This shows that the curve Cd has a nontrivial point defined over Q` .
Case 2 : f (x) has even degree. In this case we assume that f (x) has aroot modulo `. We will show that there is an integer r such that d · f (r) is
a nonzero square in Q` , and this will complete the proof. By Lemma 2.3
there exists an integer n such that ord` (f (n)) is odd and ` - f 0 (n). Writef (n) = `2k−1 s and d = `t with ` - st.
Since ` does not divide t · f 0 (n), we can choose an integer b such that
(4.1)

bt · f 0 (n) ≡ 1 − st (mod `).

Squarefree parts of polynomial values

715

Let a = `2k−1 b and r = n + a. Using a Taylor expansion we see thatf (r) = f (n + a) = f (n) + f 0 (n) · a + z · a2
for some integer z. We may write z · a2 = `4k−2 c for some c ∈ Z. Thend · f (r) = d · f (n) + f 0 (n)(ad) + d(za2 )= `2k st + f 0 (n)`2k bt + `4k−1 ct.
Letting m = st + f 0 (n)bt + `2k−1 ct we have, by (4.1),
d · f (r) = `2k · m and m ≡ st + f 0 (n)bt ≡ 1 (mod `).
Lemma 4.1 then implies that d · f (r) is a square in Q∗` .



We can now prove the local analogue of Conjecture 1.1 for polynomialsof odd degree.
Theorem 4.1. Fix N0 as in Lemma 4.2. Suppose that f (x) has odd degree.
Let p ≥ N0 be prime, and let m be an integer coprime to p. There exist
infinitely many squarefree integers d ≡ m (mod p) such that the curve Cd
has a nontrivial point over every completion of Q.
Proof. Let t be any integer such that f (t) 6= 0, and let ε ∈ {±1} be thesign of f (t). Let `1 , . . . , `r be all the primes < N0 , with `1 = 2. We definea squarefree integer δ by the formula
δ =ε·

r
Y
a

`i i ,

i=1

where ai ∈ {0, 1} is the parity of ord`i (f (t)). Note that δ·f (t) is positive andhas even valuation at every prime ` < N0 . Hence, we may write δ · f (t) =`2e` u` with u` ∈ Z not divisible by `. By Lemma 3.3 there exist infinitely
many primes q such thatq ≡ δ −1 m (mod p),
q ≡ u−12 (mod 8), andq ≡ u`i (mod `i ) for every i > 1.
Fix any such prime q, and let d = q · δ. Note that d is squarefree andd ≡ m (mod p). We claim that d · f (t) is a square in every field Q` with
` < N0 . Indeed, for each odd prime ` < N0 we haved · f (t) = q · δ · f (t) = q · `2e` · u` = `2e` (q · u` ),and by construction q · u` is a nonzero square modulo `. Hence, d · f (t) isa square in Q` by Lemma 4.1. Similarly, we haved · f (t) = q · δ · f (t) = q · 22e2 · u2 = 22e2 (q · u2 ),and by construction q · u2 ≡ 1 (mod 8). Hence, d · f (t) is a square in Q2 .
Thus, we have shown that the curve Cd has a nontrivial point defined over

716David Krumm

Q` for every prime ` < N0 . By Lemma 4.3, the same holds for all primes` ≥ N0 . Finally, Cd has a nontrivial point over R because d · f (t) is positiveand hence a square in R. Since we have infinitely many choices for q, theabove construction yields infinitely many squarefree integers d ≡ m (mod p)
such that Cd has a nontrivial point over every completion of Q.
Remark. We are not aware of any example of a polynomial f (x) of odd
degree and a prime p for which the conclusion of Theorem 4.1 fails; thisraises the question of whether the assumption that p ≥ N0 is superfluous.
For polynomials of even degree, however, a lower bound on p is necessary,as will be shown in the examples following Theorem 4.2.
We turn now to consider the local analogue of Conjecture 1.1 for polynomials of even degree. Unlike the relatively elementary proof in the caseof odd degree, the proof in this case uses results from class field theory. We
give here a brief review of the necessary definitions and theorems in orderto keep the article self-contained.
To begin we recall some basic facts about Artin symbols, as these will be
used in the proof of Lemma 4.4. See [6, p. 95] for proofs of these statements.
Let L/K be a Galois extension of number fields. Let p be a prime of OK
unramified in OL , and let P be a prime of OL lying over p. There is aunique automorphism σ ∈ Gal(L/K) such that for every α ∈ OL ,σ(α) ≡ αN (p) (mod P).
Here, N (p) is the norm of the ideal p. This map σ is denoted using the Artinsymbol ((L/K)/P). As P ranges over all the primes of OL containing p, thesymbols ((L/K)/P) form a full conjugacy class in the group Gal(L/K); thisclass is denoted by ((L/K)/p). We can now state a classical result whichis the main tool used in the proof of Lemma 4.4. We refer the reader to [6,§8.B] and [17] for further details on this theorem.
Chebotarev Density Theorem. Let C be a conjugacy class in the group
Gal(L/K). The set of primes p of OK such that p is unramified in OL and((L/K)/p) = C has Dirichlet density (#C)/[L : K].
In addition to the Chebotarev theorem we will need the following properties of the Artin symbol. These and other basic properties of the symbol
are proved in [18, Thm. 3.9].
• The prime p splits completely in L if and only if ((L/K)/p) = {1}.
• Let F/K be a Galois subextension of L/K and let P = P ∩ OF .
Then




F/K
L/K
=
.
P
P
F
For the remainder of this section we fix an algebraic closure Q and consider all number fields to be contained in Q. For any positive integer n,Squarefree parts of polynomial values

717

let ζn ∈ Q be a primitive n-th root of unity. Recall that for every integera coprime to n there is an automorphism σa ∈ Gal(Q(ζn )/Q) with theproperty that σa (ζn ) = ζna . Moreover, the map [a] 7→ σa is an isomorphism(Z/nZ)× → Gal(Q(ζn )/Q). If p is a rational prime that does not dividen, then p is unramified in Q(ζn ), so there is a well defined Artin symbolQ(ζn )/Q. One can check that this symbol is in fact equal to {σp }.
pLemma 4.4. Let h(x) ∈ Z[x] be irreducible and let L ⊂ Q be the splitting
field of h(x). Let n be a positive integer and let a be an integer coprime ton. If the map σa fixes the field F = L ∩ Q(ζn ), then there exist infinitely
many primes p such that p ≡ a (mod n) and h(x) has a root modulo p.
Proof. Let E = L · Q(ζn ) be the composite of L and Q(ζn ). By Galois
theory [8, §14.4, Prop. 21], the restriction map
Gal(E/Q) → Gal(L/Q) × Gal(Q(ζn )/Q)σ 7→ σ|L , σ|Q(ζn )is an injective group homomorphism with image
H = {(ϕ, τ ) : ϕ|F = τ |F }.
Since σa fixes F , the pair (1, σa ) belongs to H. Hence, there is an automorphism σ ∈ Gal(E/Q) such thatσ|L = 1 and σ|Q(ζn ) = σa .
Note that σ belongs to the center of Gal(E/Q) because the pair (1, σa ) liesin the center of H. Therefore, the conjugacy class of σ in Gal(E/Q) is {σ}.
From the Chebotarev Density Theorem it follows that there exist infinitely
many rational primes p, unramified in E, such that ((E/Q)/p) = {σ}. Fix
any such prime p that does not divide the leading coefficient of h(x).
Claim 1 : p ≡ a (mod n). Equivalently, the maps σp and σa are equal.
Let p be a prime of Q(ζn ) lying over p and let P be a prime of E lying overp. Thenσp =Q(ζn )/Q
p





=

E/Q
P



= σ|Q(ζn ) = σa .
Q(ζn )

Claim 2 : h(x) has a root modulo p. Let p be any prime of L lying overp, and P a prime of E lying over p. Then


L/Q
p





=

E/Q
P





= σ|L = 1, so
L

L/Q
p



= {1}.

Therefore, p splits completely in L. Let θ ∈ L be a root of h(x), and let cbe the leading coefficient of h(x). Then the minimal polynomial of θ over Q
is 1c · h(x), which has coefficients in the local ring Z(p) , so θ is integral over

718

David Krumm

Z(p) . Hence, for any prime p of L lying over p, θ belongs to the local ring
OL,p . (This can be seen using [2, Prop. 5.12].) Since p splits completely inL, the residue field of OL,p is Z/pZ. Thus, reducing the equation h(θ) = 0
modulo p we obtain a relation h(θ) = 0 over Z/pZ, and this proves theclaim.
We can now prove our main theorem for polynomials of even degree by
combining Lemma 4.4 with the following well known result.
Kronecker-Weber Theorem. For every abelian extension L of Q there
is a positive integer b such that L ⊆ Q(ζb ). Moreover, one can take b to be
the absolute value of the discriminant of L.
Proof. See [23, §V.1, Thm. 1.10] and [18, Thm. 3.3].

Theorem 4.2. Suppose that f (x) has even degree and that some irreducible
divisor of f (x) has abelian Galois group. Then there is a constant N suchthat the following holds for every prime p ≥ N : if m is any integer coprime
to p, then there exist infinitely many squarefree integers d ≡ m (mod p)
such that the curve Cd has a nontrivial point over every completion of Q.
Proof. Let t be any integer such that f (t) 6= 0 and let δ be the squarefreepart of f (t). Let h(x) ∈ Z[x] be an irreducible divisor of f (x) such that the
splitting field L ⊂ Q of h(x) has abelian Galois group. By the KroneckerWeber theorem, there is an effectively computable positive integer b suchthat L ⊆ Q(ζb ). Fix an integer N0 as in Lemma 4.2, and let N be any
integer satisfying
N > max{b, N0 , |f (t)|}.
We claim that N has the property given in the statement of the theorem.
Let p ≥ N be prime and let m be an integer coprime to p. We will showthat there are infinitely many squarefree integers d ≡ m (mod p) such that
Cd has a nontrivial point over every completion of Q. Let 2, `1 , . . . , `r be
all the primes < N , and set
α = 8p · `1 · · · `r , n = α · b.
Since ζnα is a primitive b-th root of unity, we have L ⊆ Q(ζnα ) ⊆ Q(ζn ).
Note that p does not divide 8b · `1 · · · `r since, by construction,p ≥ N > max{2, `1 , . . . , `r , b}.
Similarly, p does not divide δ since p ≥ N > |f (t)| ≥ |δ|. Hence, there exists
an integer a satisfyinga ≡ 1 (mod 8b · `1 · · · `r ) and a ≡ δ −1 m (mod p).
From the definitions it follows that a is coprime to n. We claim that themap σa ∈ Gal(Q(ζn )/Q) fixes L; in fact, it fixes the larger field Q(ζnα ).

Squarefree parts of polynomial values

719

Indeed, we haveσa (ζnα ) = σa (ζn )α = ζnaα = ζnαbecause n|α(a−1). By Lemma 4.4 there exist infinitely many primes q suchthat q ≡ a (mod n) and h(x) has a root modulo q. In particular, f (x) has
a root modulo q. Fix any such prime q ≥ N and let d = δ · q. Note thatd is squarefree and d ≡ m (mod p). By Lemma 4.3, the curve Cd has anontrivial point over every field Q` with ` ≥ N . (The only prime ` ≥ Nthat divides d is ` = q, and f (x) has a root modulo q.) Moreover, sinced · f (t) is positive, Cd also has a nontrivial point over R. By construction,q ≡ a ≡ 1 (mod 8 · `1 · · · `r ), so Lemma 4.1 implies that q is a square in
Q2 , Q`1 , . . . , Q`r . Now, d · f (t) = q · δ · f (t), and δ · f (t) is a square integer,so d · f (t) is a square in Q` for all primes ` < N . Hence, Cd has a nontrivial
point over every field Q` with ` < N . Thus, we have shown that Cd has anontrivial point over every completion of Q. Since we have infinitely many
choices for q, this construction provides infinitely many squarefree integersd ≡ m (mod p) such that Cd has a nontrivial point over every completionof Q.

Remark. All currently available evidence suggests that the Galois groupcondition included in the hypotheses of Theorem 4.2 is not necessary. However, a lower bound on p is necessary – otherwise the conclusion of thetheorem may not hold. The examples below will illustrate some of the issues that can arise at small primes.
Example. Let f (x) = x6 + 2x5 + 5x4 + 10x3 + 10x2 + 4x + 1. The prime
p = 3 is not good for f (x), because f (x) has a multiple root modulo 3(namely, x = 1). We claim that if d ≡ 2 (mod 3) is any squarefree integer,then the curve Cd does not have a nontrivial point over Q3 . In particular,
this implies that there does not exist a squarefree integer d ≡ 2 (mod 3)
such that Cd has a nontrivial point over every completion of Q. Hence, theconclusion of Theorem 4.2 does not hold if p = 3, m = 2. To prove the claim,
suppose that d is a squarefree integer such that Cd has a nontrivial point
over Q3 . We will show that d ≡ 0, 1 (mod 3). By hypothesis, there exista, b ∈ Z3 and s ∈ Q∗3 such that ds2 = f (a/b). We may assume without lossof generality that a and b are not both divisible by 3. Define a polynomial
F (x, y) ∈ Z[x, y] by
F (x, y) = y 6 · f (x/y) = x6 + 2x5 y + 5x4 y 2 + 10x3 y 3 + 10x2 y 4 + 4xy 5 + y 6 .
Letting t = sb3 , we have
F (a, b) = b6 · f (a/b) = b6 · ds2 = dt2 .
Considering the 3-adic valuation of both sides of this equation we conclude
that t ∈ Z3 . Now, allowing a and b to take all possible values modulo 9
we find that if F (a, b) ≡ 0 (mod 9), then a, b ≡ 0 (mod 3), which is a

720David Krumm

contradiction. Hence, F (a, b) is not divisible by 9, so t is not divisible by 3,
and therefore t2 ≡ 1 (mod 3). Thus,d ≡ dt2 ≡ F (a, b) (mod 3).
The conclusion that d ≡ 0, 1 (mod 3) now follows from the easily verified
fact that the map F : F3 × F3 → F3 only takes the values 0 and 1.
The above example shows that the conclusion of Theorem 4.2 can failif p is not good for f (x). The next example will show that the same canhappen if p is a good prime that is too small.
Lemma 4.5. Suppose that f (x) has even degree. Let p be a good prime
for f (x), and assume that C(Fp ) = ∅. Let d be any squarefree integer thatis a nonzero square modulo p. Then Cd does not have a nontrivial point
over Qp .
Proof. Suppose that Cd does have a nontrivial point over Qp , so that there
exist a, b ∈ Zp and s ∈ Q∗p such that ds2 = f (a/b). We may assume without
loss of generality that a and b are not both divisible by p. We consider twocases, depending on whether b is divisible by p.
Suppose first that b 6≡ 0 (mod p), so that a/b ∈ Zp . Considering thep-adic valuation of both sides of the equation ds2 = f (a/b) we see thatnecessarily s ∈ Zp . This equation then takes place in Zp , so we may reduce¯ 2 = f (x) with x, y ∈ Fp .
modulo p to obtain a solution to the equation dy
Since d is a square modulo p, we may write d¯ = α2 for some α ∈ Fp .
Then we have a point (x, αy) ∈ C(Fp ), contradicting the assumption that
C(Fp ) = ∅.
Suppose now that b ≡ 0 (mod p). Write f (x) = c2k x2k + · · · + c1 x + c0
with c2k 6= 0, and let
F (x, y) = y 2k · f (x/y) = c2k x2k + c2k−1 x2k−1 y + · · · + c1 xy 2k−1 + c0 y 2k .
We have
F (a, b) = b2k · f (a/b) = b2k ds2 = dt2 ,where t = sbk . Considering p-adic valuations we see that t ∈ Zp . The above
equation can then be reduced modulo p to obtain c2k a2k ≡ dt2 (mod p). In
particular, c2k a2k is a square modulo p. Since p does not divide a (because it
divides b), this implies that c2k is a square modulo p; thus, C(Fp ) contains
two points at infinity. Once again, this contradicts the assumption that
C(Fp ) = ∅. Since both cases have led to a contradiction, we conclude that
Cd cannot have a nontrivial point over Qp .

Example. Let f (x) = 2x8 − x6 − 8x4 − x2 + 2. The prime p = 19 isgood for f (x), and the curve C : y 2 = f (x) has no point over the field
F19 . (This and other examples of “pointless” curves over finite fields canbe found in the articles [13, 20].) It follows from Lemma 4.5 that if m is a

Squarefree parts of polynomial values

721

quadratic residue modulo 19, then there does not exist a squarefree integerd ≡ m (mod 19) such that Cd has a nontrivial point over Q19 . Hence, theconclusion of Theorem 4.2 does not hold with these values of p and m.
5. Empirical data
In this last section we summarize the results of two numerical experiments designed to test Conjecture 1.1. Unless otherwise specified, all computations were done using Sage [29].
The first experiment is a partial verification of the conjecture for a systematically chosen collection of polynomials. Given a separable nonconstantpolynomial f (x) ∈ Z[x], the conjecture states that for every large enoughprime p, the set S(f ) contains infinitely many elements from every nonzero
residue class modulo p. For purposes of checking this statement in practice,
the conjecture must be modified so that it can be reduced to a finite computation. In particular, only a finite number of primes p can be considered,and only a finite subset of S(f ) can be computed. Hence, in order to test
the conjecture we take the following steps:(1) Choose a finite set of primes, say, all primes p ≤ n for a fixed
positive integer n.
(2) Compute a finite subset of S(f ): fix a height bound B and lete ) = {S(f (r)) : r ∈ Q, f (r) 6= 0, and H(r) ≤ B}.
S(fe ) contains an element(3) For every prime p ≤ n, check whether S(ffrom every nonzero residue class modulo p.
If p is a prime for which the check in step (3) fails, we will say that p is
an exceptional prime for f (x) (although this term depends on the choice ofheight bound B). Conjecture 1.1 does not preclude the possibility that f (x)may have exceptional primes, but it does suggest that all such primes must
be relatively small. Thus, if the conjecture holds, we should not expect tofind any large exceptional primes when following the steps above. For our
first experiment we chose a large collection of polynomials and carried out
a computation which shows that, as expected, all exceptional primes forthe selected polynomials are quite small.

Proposition 5.1. Let f (x) ∈ Z[x] be a separable polynomial of degree 4,
5, 6, 7, 8 or 9, all of whose coefficients have absolute value at most 3. Thenfor every prime 19 < p < 103 , the set S(f ) contains a representative fromevery nonzero residue class modulo p.
Proof. For every polynomial f (x) satisfying the conditions of the proposition, we follow the three steps listed above taking n = 103 and B = 400.
The largest exceptional prime that occurs in the entire computation is 19.
e ) contains a representative
Hence, for every prime 19 < p < 103 , the set S(fDavid Krumm

722e ), the samefrom every nonzero residue class modulo p. Since S(f ) ⊇ S(fholds for S(f ).
Remark. The total number of polynomials satisfying the conditions of
Proposition 5.1 is 17,896.
For the second experiment we select only four polynomials, but carry out
a more extensive computation in terms of the number of primes considered.
For each of the selected polynomials f (x) we set a height bound of B = 800e ) as defined earlier. This set is then reduced modulo
and compute the set S(fp for every prime p < 10, 000, and a list of exceptional primes is kept. Onceagain, all exceptional primes that were found are very small. The precise
results of our computation are as follows:• For the fifth cyclotomic polynomial, f (x) = x4 + x3 + x2 + x + 1,
the only exceptional prime below 10, 000 is p = 5. The reduction ofe ) modulo 5 is {0, 1}.
the set S(f• For the polynomial f (x) = x7 − 3 there is no exceptional prime
below 10,000. This example illustrates a more general fact: we havenot found any polynomial f (x) of odd degree having an exceptional
prime.
• For the polynomial from Example 4, namely f (x) = x6 + 2x5 +
5x4 + 10x3 + 10x2 + 4x + 1, the only exceptional prime below 10, 000e ) modulo 3 is {0, 1}.
is p = 3. The reduction of the set S(f86• For the polynomial f (x) = 2x − x − 8x4 − x2 + 2, which appears in
Example 4, the primes 2, 3, and 97 are not good, and turn out to be
exceptional primes; the only good exceptional prime below 10, 000e ) modulo 19 consists of allis p = 19. The reduction of the set S(fnon-squares modulo 19, a result that is consistent with Example 4.
Using Magma [4] we determine that the latter two polynomials are irreducible and have nonabelian Galois group. Hence, the results of this experiment support the statement that the conclusion of Theorem 4.2 remains
valid without the assumption that some divisor of f (x) has abelian Galois
group.
We end this article by posing two questions motivated by the above
experiments.
Questions.
(1) Does the conclusion of Theorem 4.2 continue to be true if the hypothesis of an irreducible divisor with abelian Galois group is removed?
(2) For polynomials of odd degree, can Conjecture 1.1 be strengthened
to include all primes, rather than all but finitely many?

Squarefree parts of polynomial values

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Acknowledgements. The author would like to thank Dino Lorenzini for
helpful conversations during the preparation of this paper, and for valuablecomments on an earlier draft.
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David Krumm
Department of Mathematics
Colby College
Waterville, ME 04901, USA
E-mail: dkrumm@colby.edu