HMMT February

We want $a b \mid a^{2017}+b$. This gives that $a \mid b$. Therefore, we can set $b=b_{2017} a$. Substituting this gives $b_{2017} a^{2} \mid a^{2017}+b_{2017} a$, so $b_{2017} a \mid a^{2016}+b_{2017}$. Once again, we get $a \mid b_{2017}$, so we can set $b_{2017}=b_{2016} a$. Continuing this way, if we have $b_{i+1} a \mid a^{i}+b_{i+1}$, then $a \mid b_{i+1}$, so we can set $b_{i+1}=b_{i} a$ and derive $b_{i} a \mid a^{i-1}+b_{i}$. Continuing down to $i=1$, we would have $b=b_{1} a^{2017}$ so $a b_{1} \mid 1+b_{1}$. If $a \geq 3$, then $a b_{1}>1+b_{1}$ for all $b_{1} \geq 1$, so we need either $a=1$ or $a=2$. If $a=1$, then $b \mid b+1$, so $b=1$. This gives the pair $(1,1)$. If $a=2$, we need $2 b \mid b+2^{2017}$. Therefore, we get $b \mid 2^{2017}$, so we can write $b=2^{k}$ for $0 \leq k \leq 2017$. Then we need $2^{k+1} \mid 2^{k}+2^{2017}$. As $k \leq 2017$, we need $2 \mid 1+2^{2017-k}$. This can only happen is $k=2017$. This gives the pair $\left(2,2^{2017}\right).$