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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.

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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 2:12
So shall we assume it cannot be solved without elliptic curves? – quanta Apr 30 '11 at 5:30

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:


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: $$3(k-a)k(k+a)=(1-a^3)$$

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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
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 8:30
@quantumelixir, yes see… and… if you are curious where this equation comes from. – quanta Apr 19 '11 at 10:12

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