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How to solve the Diophantine equation $3(u-v)(u+v)(3u+v) = (1+2u)(1-2u+4u^2)$ over integers?

PS. I saw it here on AoPS, but could not solve it and no one has answered there.

quanta
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Amir Parvardi
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2 Answers2

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Your equation is equivalent to ${u}^{3}+3\,{u}^{2}v-9\,{v}^{2}u-3\,{v}^{3}-1=0$ which is birationally equivalent to the elliptic curve ${y}^{2}+y={x}^{3}+20$ which is rank 1 over the rationals. The obvious point (u,v) = (1,0) corresponds to the group identity, but the point (u,v) = (-2,-1) corresponds to the point (x,y) = (9/4, 41/8) which has infinite order. The torsion group has order 3.

This says nothing about integer points but still may be useful. For example, you can generate points at will on the elliptic curve and move them back to your curve to see if they are integral. I strongly suspect there are only the two integral points mentioned above.

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    To learn how to derive all the facts in Rita's 1st paragraph, you need several hours of lectures on (or reading about) elliptic curves. This is one very clever dog. – Gerry Myerson Apr 20 '11 at 02:12
  • So shall we assume it cannot be solved without elliptic curves? – quanta Apr 30 '11 at 05:30
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Here's a start:

Simplifying both sides, we get, $u^2 (u+3v)=1+9uv^2+3v^3$. Writing the RHS as a factor of $(u+3v)$, $u^2(u+3v)=1+9v^2(u+3v)-24v^3$. This yields:

$$(u+3v)^2(u-3v)=(1-24v^3)$$

and then make the substitutions $a=u+3v$, $b=u-3v$ to obtain $(a-b)^3=9(1-a^2b)$.

This means that $(1-a^2b)=3k^3$ for some $k \in \mathbb{Z}$. Multiplying $a-b=3k$ by $-a^2$ and using the previous equation, we get: $$3(k-a)k(k+a)=(1-a^3)$$

quantumelixir
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  • @quantumelixir, Brilliant! Thanks for your solution! – quanta Apr 18 '11 at 20:22
  • Thanks. Glad it helped! :) – quantumelixir Apr 18 '11 at 20:25
  • I got stuck trying to reconstruct the Cardano part: Somehow I use the fact that $a^6+\tfrac{7}{3}a^3+1$ is a square? (of course that implies $3|a$ but not sure what to do next) and that $\tfrac{1}{3}(\frac{a^3-1}{2}-\sqrt{\text{the square}})$ is a cube? – quanta Apr 18 '11 at 20:35
  • I second quanta's query about use of Cardano. While I haven't done the calculations, I'd be surprised if Cardano didn't lead to some diophantine equation at least as hard as the cubic in $k$ and $a$. – Gerry Myerson Apr 19 '11 at 01:29
  • Thanks for the solution! :) – Amir Parvardi Apr 19 '11 at 02:16
  • @quanta, @gerry-myerson: My bad; It looks like I paid the price for a hand-waving argument. You are right in pointing out that the diophantine equations that arise out of the Cardano method are much harder than the original cubic itself. I presumed that they could be solved by Cardano's method, as I was able to find the solutions on my computer. – quantumelixir Apr 19 '11 at 06:07
  • @amir-hossein: Treat this question as unsolved, until we are able to find a solution that circumvents the problem of finding the roots of a cubic. I suggest you remove the "accepted" status of my answer; I'm reverting my answer to the previous edit. I'm sure there is a better way to solving this that I'm not able to see right away. – quantumelixir Apr 19 '11 at 06:09
  • Cubic diophantine equations in two variables are hard; there are techniques that generally work, but they are not easy to master (and I have not mastered them myself). If no suitable answer is posted here, you might try asking for pointers on MathOverflow. – Gerry Myerson Apr 19 '11 at 07:10
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    Yes. This seemingly innocuous problem seems to have connections to elliptic curves (Mordell's Theorem), and I too am beginning to suspect that an elementary solution can't be found. For instance, using arithmetic modulo $m$ helps only if you want to prove non-existence of solutions to a diophantine equation; but we know this equation has at least two solutions in integers. – quantumelixir Apr 19 '11 at 08:30
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    @quantumelixir, yes see http://math.stackexchange.com/questions/30659/strange-cubic-diophantine-equations and http://math.stackexchange.com/questions/30518/where-is-the-mistake-in-this-incorrect-proof-in-eisenstein-integers if you are curious where this equation comes from. – quanta Apr 19 '11 at 10:12
  • @quanta: I think the solution to this question could use ideas from the links you have provided. Thanks for pointing it out. Will take a look at them in detail a little later. – quantumelixir Apr 19 '11 at 14:02