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.
 A: Setting $z=x+1$, we need $y^2 = z^3-z = z(z^2-1)$. Since $\gcd(z,z^2-1)=1$, we need $z=m^2$ and $z^2 - 1 = n^2$. This forces $z^2-n^2=1 \implies (z+n)(z-n) = 1 \implies z = \pm 1 \text{ and }n=0$. Apart from this, clearly $z=0$ is also a solution. Hence, the only solutions are $$(x,y) = (0,0);(-1,0);(-2,0)$$
A: If two natural numbers are relatively prime, and their product is a square, then the numbers must be squares. (Consider the prime factorizations.) Since $x+1$ and $x(x+2)$ are relatively prime, both must be squares. Let $z=x+1$; then $x(x+2)=(z-1)(z+1)=u^2$ for some natural $u.$ Clearl this is only possible if $u=0,\ $ i.e., $y=0.$
A: 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 $y^2 = 0$.
