A number theory problem by Priyanshu Mishra

If $0 \le n \le 2016$ and $a^n$ divides $b$ , then $a^{n+1}$ divides $ab$ , and hence $a^{n+1}$ divides $a^{2017} + b$ . This implies that $a^{n+1}$ divides $b$ . Since it is certainly true that $a^0=1$ divides $b$ , we deduce that $a^{2017}$ divides $b$ . Thus $b = a^{2017}c$ for some positive integer $c$ , and we now know that $ac$ divides $1+c$ .

Since $c$ divides $1+c$ (and these two numbers are coprime), we deduce that $c=1$ , and hence that $a$ divides $2$ . Thus there are $\boxed{2}$ solutions, namely $(a,b) = (1,1)$ and $(2,2^{2017})$ .