Open Contests

Answer: $(0,1)$ and $(-1,0)$.

Since $a$ and $a+1$ are coprime, $a^2$ and $a+1$ are also coprime. Similarly $b^2$ and $b-1$ are coprime. Hence the equality can hold only in the case $a+1=\pm b^2$ and $b-1=\pm a^2$, where the signs in both equations are the same.

Let both signs be pluses. Then from the first equation we get $a=b^2-1=(b-1)(b+1)$. The second equation implies $b-1=a^2$, whence $a=a^2(a^2+2)$. If $a=0$, then $b=1$, i.e. $(a,b)=(0,1)$. If $a\ne 0$, then by dividing by $a$ we get $1=a(a^2+2)$; since $a^2+2>1$, this equation does not have integer solutions.

If both signs are minuses then by multiplying by $-1$ we get $-b+1=a^2$ and $-a-1=b^2$. These are the same equations with respect to $-b$ and $-a$ which we had previously with respect to $a$ and $b$, hence the only solution is $-b=0,-a=1$ i.e. $(a,b)=(-1,0)$.