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Cubic Surfaces and Linear Recurrences
Maurice Craig

1Introduction

Ramanujan [11, p. 341] found solutions of the Diophantine equation
x3 + y 3 = z 3 + t3

(1)

for which t = ±1 and x, y, z are the nth terms in certain sequences. These sequences
form linearly independent solutions of the linear recurrence with companion polynomial(s + 1)(s2 − 83s + 1). (See [8, p. 63]).
Hirschhorn [5], [7] derives the result from a preliminary solution of (1) in binary quadratic
forms, and suggests that Ramanujan’s approach was the same. The present paper identifies
the geometrical locus of these solutions then recalls some closely allied work, mentioned inDickson’s History but uncommon in Australian libraries.

2

Quadratic forms

In [10, vol. 2, p. 266] Ramanujan observes (albeit more concisely) thatn[(mp + nq)3 − (nq + mr)3 ]or, expanded,mn(p − r)[m2 (p2 + pr + r2 ) + 3mn(p + r)q + 3n2 q 2 ],is a symmetric function of m and n when (by inspection)
p2 + pr + r2 = 3q 2 ,a condition independent of m and n. To solve this equation writepw − rω 2 = (ω − ω 2 )(aω + bω 2 )2 ,where ω 2 + ω + 1 = 0, then equate coefficients of ω and ω 2 . Expressed symmetrically, theresult is
p + 3a2 = q + 3ab = r + 3b2 = (a + b)2 .
To provide a solution of equation (1), m and n must be proportional to cubes. But theequation reduces to an identity when they are ±1. The next simplest choices include m = 8,n = 1. After removing a common factor 3, we have thenx = (3, 5, −5), y = (4, −4, 6),z = (−5, 5, 3), t = (6, −4, 4),(λ, µ, ν) serving as an abbreviation for the binary quadratic form λa2 + µab + νb2 .
The same identity reappears in “Question 441” [9, p. 326]. See also “Question 661” and“Question 681” (p. 331), the “Comments on the papers” (pp. 393, 399, 400), and [10, vol.
2, p. 4]. Hirschhorn uses this solution of equation (1) with a + b, a − 2b replacing a, b. So hisclaim [5, p. 200] to reconstruct the path that led Ramanujan to equation (1) seems beyond
doubt.

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255

However, binary quadratic forms span a vector space of dimension 3, whence x, y, z, t
above must be linearly dependent. In fact, by a mental calculation,x − z = 4(t − y).

(2)Consequently, all solutions in the parametric form above are confined to the plane given byequation (2). This plane meets the cubic surface (1) in the line x = z, t = y, and residuallyin the conic where the plane intersects the quadric surface
4(x2 + xz + z 2 ) = t2 + ty + y 2 .

(3)Now, a straightforward calculation shows that x, y, z, t above satisfy (3) anyway, whence
the solutions of (1) in this parametric form are restricted entirely to a conic. It meets thestraight line in just two points, given by a = ±b.

3Linear recurrences

In paraphrase, the recurrent solutions of equation (1) arise from specialising the values a, b
in a solution by quadratic forms, while these forms satisfy lower–degree equations having
equation (1) as just one consequence among many. Now, the statement in [11, p. 341]
(quoted in [5, p. 199]) excludes reference to quadratic forms, without whose aid the solution
itself may be hard to derive. But can we anticipate directly from the form of the original
assertion at least its dependence on equations of lesser degree? I show next how it follows
from the reducibility of the companion polynomial.
Thus, let f (s) = (1 − αs)(1 − βs)(1 − γs) = 1 − As + Bs2 − Cs3 with s an indeterminate,α, β, γ distinct and A, B, C rational. Write p0 , p1 , p2 , ... for the coefficients in the Taylor
expansion
X1/f (s) =pn sn .
(4)
Multiplying both sides by f (s) and equating coefficients of like powers of s givespn − Apn−1 + Bpn−2 − Cpn−3 = 0, n > 3,

(5)

whence the coefficients satisfy a linear recurrence relation with constant coefficients. Instead,multiply both sides by (1 − βs)(1 − γs) and equate coefficients of like powers of s in the
expanded forms to obtainpn − (β + γ)pn−1 + βγpn−2 = αn(6)pn − (γ + α)pn−1 + γαpn−2 = β n ,pn − (α + β)pn−1 + αβpn−2 = γ n .

(7)(8)

and, by symmetry,For example, if α = 0, A = 9 and B = −1 then equations (5)–(6) both reduce topn − 9pn−1 − pn−2 = 0,
while multiplying corresponding sides of equations (7)–(8) produces
p2n − 9pn pn−1 − p2n−1 = (−1)n .
Several results for the case C = αβγ = 1 also follow easily. Suppose first that f (s) isirreducible over the rational numbers. Multiply corresponding sides of equations (6)–(8) toproduceY[ρpn − ρ(A − ρ)pn−1 + pn−2 ] = 1,(9)256

Maurice Craig

the three factors being for ρ = α, β, γ. Hence (pn , pn−1 , pn−2 ) is a point on a cubic surface(now in Cartesian coordinates) defined over the rational numbers.
If, instead, f (s) is the product of the linear factor (1 − s) with an irreducible quadratic
then we may take α = βγ = 1. In place of equation (9) there are two equations with rationalcoefficients, both formally expressible as (9), but with either a single factor, for ρ = α, orwith two factors, for ρ = β, γ. Thus, if f (s) = (1 − s)(1 + 83s + s2 ) we findpn + 83pn−1 + pn−2 = 1,(10)(pn − pn−1 )2 + 83(pn − pn−1 )(pn−1 − pn−2 ) + (pn−1 − pn−2 )2 = 1.

(11)

The point with coordinates (pn , pn−1 , pn−2 ) lies on the conic determined by the common
solutions of equations (10), (11). No third relation among the three coordinates can bealgebraically independent of these two, because three independent equations would have
only finitely many solutions.
Equation (4) implies thatX(λ + µs + νs2 )/f (s) =An sn ,An = λpn + µpn−1 + νpn−2 .
All such sequences An solve the one linear recurrence (5). Moreover, in a relation such as (6)or (9) we can eliminate (pn , pn−1 , pn−2 ) in favour of (An , Bn , Cn ), where the components ofthis vector come from three linearly independent triples (λ, µ, ν). Conversely, the coordinatesof any such points (An , Bn , Cn ) can be expressed linearly in terms of pn , pn−1 , pn−2 , whence
all these points lie on a conic.
This conclusion is not peculiar to solutions of equation (1), but valid for all sets ofthree independent solutions of a cubic recurrence with reducible companion polynomial.
Nevertheless, in the Appendix I carry out the calculations for the example above, thenapply them to deduce Ramanujan’s result explicitly from equations of lesser degree.
Thus we see again (here from reducibility, as above from a dimension argument involving
quadratic forms) that Ramanujan’s identity (1) can be only one member, by no means ofleast degree, in a family of identities satisfied by the nth terms of the same sequences. The
relation of least degree for the sequences in [11, p. 341], namely an + 4bn − 4cn = (−1)n ,can be verified numerically for the first seven triples, given in [6]. This relation follows fromthe identity(1 + 53s + 9s2 ) + 4(2 − 26s − 12s2 ) − 4(2 + 8s − 10s2 ) = 1 − 83s + s2 ,on dividing both sides by (1 + s) times the right member. Cf. equation (2).

4

A cognate identity

As a second application of the theory developed above, I note a further result concerning
recurrence–derived points on a cubic surface. Thus, subtracting (8) from (7) yields(β − γ)(pn−1 − αpn−2 ) = β n − γ n ,two more relations following by symmetry. Hence by multiplying corresponding sides,
pn−1 3 f (pn−2 /pn−1 ) = R(α, β, γ)
where [1, p. 111ff.] the right side is the bialternant |α0 β n γ 2n |/|α0 β 1 γ 2 |. If now we write
Xg(s) = (1 − s/α)(1 − s/β)(1 − s/γ), 1/g(s) =qn sn
then, by the same reasoning, qn−1 3 g(qn−2 /qn−1 ) will be
R(1/α, 1/β, 1/γ) = (αβγ)2−2n R(α, β, γ).

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Moreover, we have (αβγ)g(s) = −s3 f (1/s). On eliminating R and replacing n by n + 1 wegetpn 3 f (pn−1 /pn ) + C 2n−1 qn−1 3 f (qn /qn−1 ) = 0.
For a self–reciprocal cubic f (s) = s3 f (1/s) the result is vacuous. But if f (s) = 2−(1+s)3
we see that (x, y, z, t) = (an , bn , cn , dn ) = aT is a point of the surface
x3 + y 3 = 2(z 3 + t3 ),(12)an , bn , cn , dn being the respective coefficients of sn in the power–series expansions of thefunctions(1 + s)/f (s), (1 + s)/g(s), 1/f (s), s/g(s).
This result, chronicled in [4, p. 562], is proved in [2, p. 282–285]. Might Ramanujan
have seen it? The first four solution vectors aT are (1, 1, 1, 0), (4, −2, 3, 1), (15, 3, 12, −3)and (58, −2, 46, 6). They form the rows of a matrix with determinant 36. So the points aare non–coplanar. Again, as the first 10 matrices aaT are linearly independent, these points
do not all lie on a quadric surface. Both a9 and c9 have 6 digits, the relevant determinant
being a number with 30 digits. I have not attempted to ascertain whether the first 20 points
might lie also on a second cubic surface besides (12).
In [3] I used a parametric solution of equation (12), though differently derived, to constructquadratic number fields with unusual ideal class groups.
Appendix
An = pn − 53pn−1 + 9pn−2
Bn = 2pn + 26pn−1 − 12pn−2
Cn = 2pn − 8pn−1 − 10pn−2
and let Un , Vn denote the left sides of equations (10), (11). Then equation (1) has thesolution (x, y, z, t) = (An , Bn , Cn , 1).
In fact, Bn − Cn = 2(17pn−1 − pn−2 ), Un − An = 8(17pn−1 − pn−2 ) = 4(Bn − Cn )(cf. equation (2)). To verify Bn 3 − Cn 3 = Un 3 − An 3 it is then sufficient to check thatBn 2 + Bn Cn + Cn 2 , or (Bn − Cn )2 + 3Bn Cn , is four times (Un − An )2 + 3Un An . Thiscondition reduces to pn pn−1 + pn−1 pn−2 + pn−2 pn + 82pn−1 2 = 0. By a similar reduction,
so too does the equation Un 2 = Vn , whence the truth of both equations is immediate from(10), (11).
Ramanujan’s result follows on writing an = (−1)n An etc. Varying the sign (use of f (s)instead of f (−s)) allowed us to express all our solutions of (1) with t = 1 and hence interpret(x, y, z) as Cartesian coordinates. Ramanujan avoids this simplification, presumably tosecure sequences of positive terms.
Finally, I reiterate that the purpose here is not to offer an alternative derivation but
merely to confirm that, among many similar relations, the result follows from simpler, more
fundamental equations.

References[1] A.C. Aitken, Determinants and Matrices (Oliver and Boyd/Interscience Publishers 1954).
[2] L. Chanzy, Solution (to a problem of R. Amsler), Nouv. Ann. Math. 16 (1916), 282–285.
[3] M. Craig, A type of class group for imaginary quadratic fields, Acta Arith. XXII (1973), 449–459.
[4] L.E. Dickson, History of the Theory of Numbers, vol. 2 (Chelsea New York 1966).
[5] M.D. Hirschhorn, An amazing identity of Ramanujan, Math. Mag. 68 (1995), 199–201.

258[6] M.D. Hirschhorn, A proof in the spirit of Zeilberger of an amazing identity of Ramanujan, Math. Mag.
69 (1996), 267–269.
[7] M.D. Hirschhorn, Ramanujan and Fermat’s Last theorem, AustMS Gazette 31 (2004), 256–257.
[8] D.J. Lewis, Diophantine Equations: p–adic Methods, in: W. J. LeVeque, ed., Studies in Number Theory(Prentice Hall, Inc. New Jersey 1969).
[9] S. Ramanujan, Collected Papers (AMS Chelsea Publishing 2000).
[10] S. Ramanujan, Notebooks (2 volumes) (Tata Institute of Fundamental Research Bombay 1957).
[11] S. Ramanujan, The Lost Notebook and other Unpublished Papers (Narosa New Delhi 1988).
BHP Billiton Exploration and Mining Technology, G.P.O. Box 86A, Melbourne VIC 3001
Received 25 November 2004, accepted 18 March 2005.