# Third degree Diophantine equation

I saw an exercise which was meant to find integers all integers $m,n$ satisfying $2m^2 + 5n^2 = 11(mn-11)$. I found them using the factorization $(m-5n)(2m-n)=-11\cdot 11$. However, what kind of methods there are to solve the original problem, find all $(m,n)\in\mathbb{Z}\times\mathbb{Z}$ satisfying $2m^2+5n^3=11(mn-11)$? I haven't solved many cubic Diophantine equations so I was just wondering if there is some birational transformation to convert the equation to a Weierstrass form of an elliptic curve.

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The equation $2y^2+5x^3=11(xy−11)$ describes an elliptic curve. Given any elliptic curve, you can perform a change of variables to put it into Weierstrass form, and if the field of definition has characteristic different from 2 or 3, you can even put it into the form $y^2 = x^3 + ax + b$. In this particular case, to find the transformation is easy: first scale $x$ and $y$ to get rid of make the coefficients of $x^3$ and of $y^2$ equal to 1. Then write $y' = y + \alpha x$ to get rid of the $xy$-term. This will introduce an $x^2$ term. So now, translate $x$ to get rid of the $x^2$ term (I haven't actually done the computation).
An important thing to note is that while the notion of rational points does not depend on the model over $\mathbb{Q}$ that you are working with (as long as you only change variables over $\mathbb{Q}$), the notion of integral points does depend on the integral model. A theorem of Siegel says that on any Weierstrass model of an elliptic curve, there are only finitely many integral points, and there are bounds on their height in terms of the coefficients of the model. But as far as I know, to actually find the points comes down to brute force methods: e.g. if you manage to compute generators of the Mordell-Weil group, i.e. of the group of the rational points, and the torsion subgroup, then you just check all possible linear combinations up to the given bound. There are better methods using elliptic logarithms, but they are theoretically more involved.
Edit: some more info on your concrete curve: To get a Weierstrass model you can replace $y$ by $y'/20$ and $x$ by $-x'/10$ in the original equation, resulting in $y'^2+11x'y' = x'^3 - 2^35^211^2$, confirming your hint. The elliptic curve has rank 1, as you say, and no torsion. The point $P=(22,-88)$ generates all the rational points on the curve under the group law. Integral solutions of the original equation correspond to points on the Weierstrass model with integral $x$-coordinate divisible by 10 and integral $y$-coordinate divisible by 20 (see our transformation). So the naive way of ruling out integral solutions to the original equation is to check all multiples of the generator $P=(22,-88)$ under the group law up to the given bound and convince yourself that no point satisfies the divisibility criteria. However, the bound of Baker, referred to in the wikipedia article, is huge and the computation might not actually be feasible. A possibly more promising approach is to write down the polynomial that computes $Q\mapsto Q\oplus P$ and see whether this operation always introduces higher and higher denominators that never cancel the numerators.
I got some hint that the Weierstrass form of the curve is $y^2 + 11xy = x^3 - 24200$ and its rank is one with generator $(x, y) = (22, -88)$. But at least Silverman's book has no index word generator of an elliptic curve so how to define a generator? And how this proves that there is no integer points of the original equation? – Jaska Dec 12 '10 at 12:20