On integer solutions of $3(u-v)(u+v)(3u+v) = (1+2u)(1-2u+4u^2)$ 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.
 A: 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.
A: 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: $$3(k-a)k(k+a)=(1-a^3)$$
