2011 China Western Mathematical Olympiad

The solution pairs consist of $(0,0)$ and $(-1,-1)$.

If one of $a$ and $b$ is $0$, it is obvious that the other is also $0$.

Now we assume that $ab\ne 0$, select a large prime $p$ such that $p>|a+b^2|$, it follows from Fermat's little theorem that $$a^p+b^{p+1}\equiv a+b^2(\mathrm{mod}p),$$ As $p\mid (a^p+b^{p+1})$ and $p>|a+b^2|$, we have $a+b^2=0$.

Then we select another prime $q$ such that $q>|b+1|$ and $(q,b)=1$. Let $n=2q$, then we have $$a^n+b^{n+1}=(-b^2{)}^{2q}+b^{2q+1}=b^{4q}+b^{2q+1}=b^{2q+1}(b^{2q-1}+1).$$ It follows from $$n\mid (a^n+b^{n+1})\text{and}(q,b)=1,\text{ that }q\mid (b^{2q-1}+1).$$ As $b^{2q-1}+1\equiv (b^{q-1}{)}^2\cdot b+1\equiv b+1(\mathrm{mod}q)$, and $q>|b+1|$, it follows that $b+1=0$, i.e. $b=-1$, and so $a=-b^2=-1$.

In conclusion, there are only two solution pairs $(0,0)$ and $(-1,-1)$.