<s>[INST] How to find all pairs $(x,y)$ of integers such that $y^2 = x(x+1)(x+2)$? Here $y^2$ is divisible by $12.$ And satisfying all those conditions I think $y=0$ is the only solution. But I can't show it mathematically.
 [/INST] Assume $y \ne 0$.
If $m | y^2$ $m>2$ then $m$ divides exactly 1 of $x, x + 1, x + 2$ so $m^2$ divides exactly 1 of $x, x+1, x +2$ so each $x, x+1, x + 2 = 2^nm^2$ for some (maybe 0) power of two and some $m$ odd.
Suppose $m, n$ odd and $n^2 = m^2 + 2$.
Then $(n + m)(n - m) =2$. But this has no integer solution.
So $x$ and $x+2$ aren't both odd.  So $x$ and $x+2$ are both even.  $2$ divides one of them and $4$ divides the other.  So $8|y^2$ so $16|y^2$ and $8$ divides the other.  So we have:
$8*4^l*m^2 = 2*k^2 \pm 2$ for odd $m,k$ so  $4^{l+1}m^2 = k^2 \pm 1$ which means $2^{l+1}m = \pm \sqrt{k^2 \pm 1}$ but that can only happen if one of $x$ or $x+2$ is 0.
So one of $x, x+1, x+1$ is $0$ and $y^2 = 0$.
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