AustriaMO2011

For brevity, we name $n=a^b+b$ and $m=a^{2b}+2b$.

For $a=b=0$, we obtain $n=m=1$, and therefore $n|m$. We see that $(0,0)$ is a solution.

For $a=0$ and $b>0$, we obtain $n=b$ and $m=2b$, and again $n|m$. We see that $(0,b)$ is, in fact a solution for all values of $b$.

For $a=1$, we obtain $n=b+1$ and $m=2b+1$. The only way $(b+1)|(2b+1)$ can hold is for $b=0$. We see that $(1,0)$ is also a solution.

For $b=0$, we obtain $n=1$ and $m=1$, and again $n|m$. $(a,0)$ is therefore a solution for all values of $a$.

We now consider the general case, where $a>1$ and $b>0$. Since $$a^b+b\mid a^{2b}+2b=(a^b{)}^2-b^2+(b^2+2b)=(a^b+b)(a^b-b)+b(b+2),$$ we see that $a^b+b\mid b(b+2)$ must hold. For $b=1$, we must have $a+1|3$, and therefore $a=2$. The only solution in this case is therefore $(2,1)$.

We now consider large values of $a$, specifically $a\ge 3$. For $a=3$, we have $n=3^b+b$, and we can show that $3^b+b>b(b+2)⟺3^b>b(b+1)$, which yields a contradiction. We show this by induction. For $b=1$, we have $3>1\cdot 2$, which is certainly true. If we can now assume that $3^k>k(k+1)$ is true, and wish to show $3^{k+1}>(k+1)((k+1)+1)$, we note that the latter expression results from the former by multiplying the left side with $3$ and the right side with $\frac{k+2}{k}$, which is certainly not greater than $3$, which proves the induction. If $a>3$, we have $n>3^b+b$, which is then also certainly impossible.

Finally, the case $a=2$ remains. As before, we can show that $2^b>b(b+1)$ for $b>5$ (noting $32>30$). The only possible values for $b$ remaining are therefore $2$, $3$ and $4$. For $b=2$, we would have $6|8$, for $b=3$ we would have $11|21$, and for $b=4$ we would have $20|24$, none of which is true.

In summary, the solutions are $(2,1)$, all pairs $(a,0)$ and all pairs $(0,b)$. qed