An algebra problem by Mohammed Imran

Let $u=xy+2$ and $v=x+y+1$ . Substituting in, the equation becomes $2u^2-6(v-1)^2=(v-2)^3-6$ .

Expanding both sides, $2u^2-6v^2+12v-6=v^3-6v^2+12v-8-6$ . Cancelling terms, we're left with $2u^2=v^3-8$ .

Clearly, $v$ must be even; put $v=2b$ to get $2u^2=8b^3-8$ , or $u^2=4b^3-4$ .

Again, $u$ must be even; put $u=2a$ and divide through by $4$ to get $a^2=b^3-1$

This is an example of Mordell's equation. It can be shown* that this equation has exactly one solution in integers, namely $a=0$ , $b=1$ . Substituting back through the above, we find $xy=-2$ and $x+y=1$ , so that the only solutions are $(x,y)=(-2,1)$ and $(x,y)=(1,-2)$ , ie there is exactly $\boxed1$ solution to the original equation ignoring ordering.

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* proofs of this can be found here (using factorisation over Gaussian integers) or here (using elementary number theory). Alternatively, the existence of a unique integer solution is a consequence of Catalan's conjecture.