How to solve a binary cubic equation in integers

Abstract Given any polynomial in two variables of degree at most three with rational integer coefficients, we obtain a new search bound to decide effectively if it has a zero with rational integer coefficients. On the way we encounter a natural problem of estimating singular points. We solve it using elementary invariant theory but an optimal solution would seem to be far from easy even using the full power of the standard Height Machine.


Introduction
This paper is a kind of sequel to [7], where a sharp "search bound" for rational solutions of a quadratic equation in several variables was obtained.Namely if the quadratic equation has a rational solution, then there is one whose height is bounded above in terms of the height of the quadratic.Here we prove the following analogue.THEOREM.For H ≥ 1 let F(X, Y) in Z[X, Y] be a polynomial of degree at most three, with coefficients of absolute values at most H, such that the equation F(x, y) = 0 has a solution in integers x, y.Then there is a solution with max{|x|, |y|} ≤ exp (20H) 600000 .(1) To this day we know no analogue at all of such a result for rational solutions, even for Mordell's y 2 = x 3 + m.And similarly for integral solutions of binary quartics, even for x 4 − 2y 4 + xy + x = m.
The result may not surprise the expert, who will recognise the usual bad dependence on H.But in the way we have stated the result, it would be false with the exponent 600000 replaced by any κ < 1 (even if we allow larger constants).So while it is by no means sharp, it is at least not too blunt.
The impossibility of κ < 1 arises in special situations only when F(X, Y) is not absolutely irreducible or of degree less than three.This has been shown already by Kornhauser [6].
When F is absolutely irreducible and the genus is one, then Baker and Coates [1] had obtained an upper bound for the size of all integer solutions (even with F of arbitrary degree), which in our case reduces to exp exp exp (2H) 10 59049  .
This was improved to a single exponential by Schmidt [14], which in our case reduces to something like (1) with 600000 replaced by 12 13 = 106993205379072 but no explicit multiplying constant.These authors used the Riemann-Roch Theorem to reduce to linear forms in logarithms.Sometimes Runge's Method secures a polynomial bound, which then can be comparatively small.When F is absolutely irreducible and the genus is zero, Poulakis has carried out a detailed study in [9] and [10] (and see [8] for rational points).As in [1] he treated equations of arbitrary degree, and in [10] even over an arbitrary number field, so that it is hardly a surprise that for cubics over Q our own bounds are numerically better.Later on we shall refer to some of his results.
In this genus zero situation there may be infinitely many solutions and in that case we have to be content with a search bound in the sense above.This aspect may be slightly less familiar to the expert.Also the size of the singular point, while not at all crucial for our result, seems to raise new questions related to the Height Machine in diophantine geometry and involving the classical algebraic geometry of cubics.
Thanks to the abc conjecture one expects polynomial bounds for y 2 = x 3 + m.Also y 2 = x 3 + kx + m is covered by similar conjectures (see for example Silverman [15, p.268]).Maybe there is always a polynomial bound when the number of solutions is finite.
Our methods here are rather standard, except perhaps for the invariant theory to get at singular points.Also we try to reduce the cubic to a particular "semi-rectangular" form whose highest homogeneous part is XY(X + Y).This allows a subsequent reduction to Tate form using explicit constructions without the need for Riemann-Roch (but sometimes the formulae involved are quite complicated).In special cases it also allows the solutions to be found by ad hoc elementary methods without the need for Runge.We have chosen to present in detail the worst estimate (based on linear forms in logarithms), and then for brevity omit the details when it is reasonably clear that subsequent estimates are much better.But later we also give the details for a slightly different use of linear forms in logarithms, and then again we relax.And we usually give the details in the more elementary estimates, even though they are comparatively infinitesimal especially our best estimate (which is even sharp), because these occupy relatively little space.
Throughout we use the notation to begin with for algebraic coefficients.

Preliminaries on singular points
We use h(α) or more generally h(α 1 , . . ., α n ) for the (non-homogeneous) logarithmic absolute height of algebraic numbers, with H = e h (non-logarithmic).LEMMA 1. Suppose F is absolutely irreducible with coefficients in a number field K such that H(a, b, c, d, e, f , g, k, l, m) ≤ H.
If the corresponding curve has genus zero with a singular point (x, ỹ) then x, ỹ are in K and H(x, ỹ) ≤ 500H 5/3 .
Proof.There is a discriminant b, c, d, e, f , g, k, l, m] (actually primitive), which vanishes precisely at the singular cubic curves (it is essentially Ruppert's R of [11, p.181], the Macaulay resultant of F, ∂F/∂X, ∂F/∂Y).It has 2040 terms (we checked this number given in [5, p.4]), but it can be calculated most easily by where are the standard invariants for example in [4, p.160] or [12, pp 189-192].Here the "dictionary" identification in [4] between the Dolgachev notation and the Salmon notation has to be extended to our notation (2); namely both sides have to be identified with (a, b, e, c, f , k, d, g, l, m) when we identify the variables (T 0 , T 1 , T 2 ) in [4] with our (X, Y, Z).
When the singular point is a node, it can be shown that the gradient evaluated at the coefficient vector does not vanish.One way of seeing this would be by [5,  ( As G has degree 12, the projective height in (4) is of order at most H 11 and that in ( 5) is exactly H(x, ỹ) 3 , so we obtain the exponent 11/3 instead of 5/3.However this approach fails for a cusp, because then all the coordinates in (4) vanish, thanks to the defining equations T = S = 0 for cuspidal cubics.
We present an alternative argument giving at first 11/3 that can then be modified to yield 5/3.
If further, F 1 ( P1 ) = 0 (so that P1 lies on the curve C 1 corresponding to F 1 ), then (so that P lies on the curve C corresponding to F -this curve is of course the image of C 1 under the map corresponding to ).
And if yet further, ) [3] (so that P is singular on C).Now any nodal F can be transformed into say 2 with node P1 = (1, 0, 1).After some computation we find that (8) holds with λ 1 = −64.Thus the node P = P1 satisfies (9); and this is nothing other than the projective equality between (4) and ( 5) above.
As mentioned, this gives the exponent 11/3 in the present lemma.But we can repeat the whole argument with G replaced by T (and φ 12 by φ 6 ).Unfortunately (8) no longer holds at a node, but it does at a cusp P1 = (0, 0, 1) with 1 and λ 1 = 4/27.In this case we can proceed to an explicit bound as follows.
We find that 5832T has degree 6 and integer coefficients with absolute values at most M = 5832 and with N = 103 terms.On differentiating the degree goes down to 5 and then we pick up some extra factors at most 6.We find by standard methods that H(x, ỹ) 3 ≤ 6MNH 5 so that H(x, ỹ) ≤ 154H 5/3   for a cusp.
To get a similar bound for a node, we note that as T 2 + 64S 3 = 0 for T = T(F), S = S(F) not both zero, there is a (unique) U with S = −U 2 , T = 8U 3 .Now so that projectively we have ∇T + 12U∇S (instead of the ∇T for a cusp).This time we need to know that 11664S has degree Now the height estimation is not quite so standard, and we have to return to first principles with valuations.We find and so this time That completes the proof of Lemma 1.
Since the invariant T worked for cusps, one might try the lower degree S instead for nodes.But unfortunately the key equation ( 8) fails for both nodes and cusps.Using different methods on S we can go a bit further and obtain the exponent 3/2 < 5/3 for cusps; but we omit the details because the argument fails for nodes.
We believe that the curious exponent 5/3 can be reduced, perhaps even to any σ > 1.This would be best possible in view of the family (10) with (x, ỹ) = (n, n 2 ).It seems that in (4) there are in some sense large common factors corresponding to reducibility or cusps.
For example, let us drop from projective P 9 to affine A 2 using a two-dimensional family linearly parametrized by F = F 1 + uF u + vF v with fixed F 1 , F u , F v chosen "at random".Then G = 0 defines a curve in A 2 of degree 12 (seemingly absolutely irreducible of genus 10), and the various ∂G/∂t in (4) are polynomials in u, v of degree at most 11.The divisor of a typical such polynomial is liable to be D 132 − 11∞ 12 , where D 132 is a sum of 132 different points and ∞ 12 is the sum of the 12 points at infinity.However it appears that for our ∂G/∂t we get the special form where i, j are determined by F containing the term tX i Y j , and where again the subscripts denote the number of different points.From computation it is practically certain that D 21 and D 24 correspond to reducible F and cuspidal F respectively (but D X 6 , D Y 6 , D 6 probably have no natural geometric interpretation).
We may note in passing that while the cuspidal F are defined by T = S = 0, the reducible F seem to be defined only as a determinantal variety and thus far from a complete intersection -see [11, p.178], where no fewer than 45 equations are implicit.
If indeed (11) holds, then we find that the degree of ∇G in P 9 (C( )) is only 18 due to the cancelling of D 21 and D 24 .So by the Height Machine, provided F is not reducible or cuspidal, we would get from ( 5) and ( 4) where h 1 is the height of (u, v) with respect to a divisor of degree 1.As u, v are themselves of degree 12 we have h , and also h(u, v) ≤ log H + O(1); so we end up with In fact this gives any σ > 1/2.It does not contradict the example (10) because that is not linearly parametrised.
Of course dropping to A 2 , and a special one at that, loses a lot.Still, it is tempting to think that something like (11) may sometimes persist in P 9 or at least on some open subset of the singular locus G = 0.But even so there is no suitable Height Machine.
Maybe it is relevant here to recall dimensions, some of which are clear from the above discussion.Our cubics live in nine dimensions, and the nodal cubics occupy eight of them (with degree 12).Both the cuspidal and the reducible cubics make up seven dimensions (with degrees 24 and 21 respectively), and the unions of three lines come down to six (with degree 6).
There is a natural analogue of Lemma 1 for quadratic F that are not squares.Then using the determinant instead of G we obtain rather quickly the bound H(x, ỹ) ≤ √ 5H (no trouble with cusps).Here the exponent is sharp, as the family (Y − nX)(X − n), based on the same principle as (10), shows.So we will assume |x| > 12H 2 .

Preliminaries on special equations
The following argument was found by examining the Taylor expansions at ∞.We have and so |x| 2 ≤ 6H|y|.Now and so Thus x 2 + gy + l = 0; and then also kx + m = 0 contradicting |x| > 12H 2 .This settles (a).
Observe that if k = m = 0 then there are infinitely many solutions with y = 0 (due to reducibility).
Observe that if f = 0 then for example l = 1 yields infinitely many solutions with y = −x 3 − m.And if l 3 − f 3 m = 0 then f = 1 yields infinitely many solutions with x = −l (also reducibility).
Next, when F has degree three and is not absolutely irreducible but irreducible over Q, such as X 3 − 2Y 3 or X 3 − 2, then it is a product of three conjugate polynomials of degree one.Then a, d are not both zero and after permuting we can assume a = 0.So F = aLL L for L = X + βY + γ and its conjugates.Then β is not in Q, otherwise F would have no zeroes even in Q 2 .
Using the notation H generally for the height of a polynomial via its vector of (algebraic) coefficients, we can check that for any L 1 , L 2 , L 3 in Q[X, Y] of total degree at most one (in the usual way via Gauss's Lemma and well-known archimedean inequalities in particular [13, equation (41) of corollary 12, p.249].
For our quantity H we have and so H(L) ≤ 4H 1/3 .Now using and the same bound for |y|.Thus max{|x|, |y|} ≤ 32H 2/3 (15) for all solutions.This is by far the best of all our estimates.It is amusing to note that again the idea of (10) gives the family (y − nx) 3 = 2(x − n) 3 showing here too that the exponent is sharp (probably the only such search bound in this paper).Note that if F = 0 does have a rational solution, then the corresponding curve splits into three lines meeting at the corresponding point.
And if F is reducible over Q (still of degree three), then it is LQ for linear L and quadratic Q over Z. Now again from [13] H ≥ H(LQ) ≥ 2 −6 H(L)H(Q) so [6] for Q (and even L) gives the search bound max{|x|, |y|} ≤ (896H) 320H .
From now on we shall assume that F is absolutely irreducible (so we can speak of genus) of degree three.

Genus one
As previously indicated, the search bounds obtained in this section are for all solutions.We first treat the case of three points at infinity.In geometric language we construct a sequence of rational maps where the curve C is defined by F = 0.But we shall not be pedantic about this.
Assume for simplicity that a, d are not both zero (we will address this issue later).We can assume a = 0, and then we factorise with L = Q(α, α , α ) of degree D ≤ 6.We increase the symmetry by transforming this to X 1 Y 1 (X 1 + Y 1 ) and then we kill the coefficients of We note that the three pairs cannot be (0,0) otherwise F 1 would be reducible.The transformation is given by And in the other direction we have where 0 = a 4 δ 2 is (up to sign) the discriminant of F 0 .Here the coefficients of X 1 , Y 1 are in O L because aα, aα , aα and aαα , aα α , aα α are.So are ξ , η because they are polynomials over Z in e, f , g, a, α, α , α where each α i α i α i comes multiplied by a I for I = max{i, i , i } (we shall repeatedly use this property in what follows).Thus Next we transform (16) to Tate form.We used Riemann-Roch in the standard way, but the reader can be spared this through the identity with and inverse (a map C 2 → C 1 between curves as above) where Finally we pass to Weierstrass form by means of and Here From all the above it should be clear that e 3 , k 3 , m 3 lie in L. But in fact they lie in K = Q(α).This can be seen by direct computation.But here is a slicker method.
If L = K it is trivial.Otherwise L is a quadratic extension of K with a galois generator that interchanges α , α (so changes the sign of δ).Extending this to L(X, Y) we see from ( 18) and (19) that it also interchanges X 1 , Y 1 .So by (16) it interchanges k 1 , l 1 ; but fixes f 1 and m 1 (here we implicitly used the algebraic independence of X 1 , Y 1 ).Now we get what we want from (28).
As f 1 , k 1 , l 1 , m 1 are in O L so are e 3 , k 3 , m 3 and even in Going further along these galois lines, we see from (23) that X 2 is fixed but Y 2 is sent to Also X 3 is fixed, and a short calculation then shows that Y 3 is sent to 2F 1 (X 1 , Y 1 ) − Y 3 .Let x, y in Z satisfy F(x, y) = 0, and define (x 3 , y 3 ) through (x 1 , y 1 ) and (x 2 , y 2 ) according to (18), ( 19), ( 23), (26).We deduce from the above that x 3 lies in K, and, because F 1 (x 1 , y 1 ) = 0, that y 3 is sent to −y 3 .Thus for example a 2 δy 3 is in K.And so x 3 , a 2 δy 3 are in O K .As (a 2 δy 3 ) 2 = 0 (4x 3  3 + e 3 Thus with C/2 = 36 51516 54 216 .We now proceed to estimate h 3 . It can be checked that f 1 , k 1 , l 1 , m 1 are polynomials of total degree at most 9 in a, e, f , g, k, l, m and of total degree at most 12 in α, α , α .By (14) we have Next we check that f 1 , k 1 , l 1 , m 1 each involve at most N = 574 terms (who says mathematicians can't count) and their coefficients in Z have absolute values at most M = 92.It follows that We now have to make our way back from x 3 , y 3 to x, y.This is relatively easy (and the exponent κ = 559764 will not change except in a silly way right at the end).
First (27) gives which by (31) and ( 25) is at most 3C H κ .Now we have to be careful of the denominators in (24).Assume for the moment that y 2 = 0. Then from (23) we see that x 2 = −x 1 y 1 = k 1 , and so (24) gives 16).Thus by the first in (17) y 1 = −m 1 /l 1 .This leads to It follows that for x 1 , y 1 defined in terms of x, y by (35), (36) that μ 1 = max{|x 1 |, |y 1 |} ≤ 154H 4  1 for for α in Q.Here ν = α 2 e + αf + g = 0 otherwise there would be a singularity at infinity ∞ α .Now it suffices to put with Now (38) is none other than the Tate form, and the reader can easily believe that the resulting estimate is comparable with (33).

Genus zero and singular point infinite
Now F 0 is a(X − αY) 2 (X − α Y) or a(X − αY) 3 as above, with singular point ∞ α as above.
In the first case we reduce to (34), and with the Runge-type Lemma 3(a) again we end up with (33) for all solutions.A polynomial bound can also be found in [10, theorem 1•2(i), p.252].
If β = 0 then there can be infinitely many solutions; for example y = x 3 , so only a search bound is possible.Now F = a(X − αY) 3 + e(X − αY) 2

LEMMA 2 .
Let K be a number field of degree D over Q with ring of integers O and discriminant .Suppose x, y are in O with y 2 = ax 3 + ex 2 + kx + m for a, e, k, m in O and non-zero discriminant on the right.Then max{h(x), h(y)} ≤ (12D) 17172D | | 216 exp (4050Dh ), where h = h(a, e, k, m).Proof.This follows at once from Theorem 2•2 of Bérczes, Evertse and Györy [2] (p.730) with n = 3 and S as the set of all infinite places.Of course it is linear forms in logarithms which are responsible for the large bound in Lemma 2. By contrast the next result uses (concealed) Runge-type methods so the bounds are much smaller.LEMMA 3. (a) Suppose x, y are in Z with x 2 y + gy 2 + kx + ly + m = 0 (12) for g, k, l, m in Z of absolute values at most H ≥ 1 with not both k, m zero.Then max{|x|, |y|} ≤ 145H 4 .(b) Suppose x, y are in Z with x 3 + fxy + ly + m = 0 for f, l, m in Z of absolute values at most H ≥ 1 with f and l 3 − f 3 m non-zero.Then max{|x|, |y|} ≤ 15H 6 .Proof.(a) If y = 0 then kx + m = 0 and we get the bound H.So from now on we assume y = 0.If |x| ≤ 12H 2 then we use (12) to see that y divides kx + m.If kx + m = 0 this implies |y| ≤ |kx + m| ≤ 13H 3 and we are done.If kx + m = 0 and g = 0 then |y| ≤ |gy| = |x 2 + l| ≤ 145H 4 , while if g = 0 then x 2 + l = 0 contradicting |x| ≤ 12H 2 .