<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] 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 $$(x,y) = (0,0);(-1,0);(-2,0)$$
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