Diophantine Equation problem: Find all pairs of integers (a,b) such that $ab|a^{2017} + b$ [closed]

Find all pairs of positive integers $(a,b)$ such that $a^{2017} + b$ is a multiple of $ab$, i.e. $ab | a^{2017} + b$.

Solutions using casework are most appreciated, already tried infinite descent method

closed as off-topic by Namaste, Xam, Davide Giraudo, C. Falcon, ShaileshJun 3 '17 at 0:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Xam, Davide Giraudo, C. Falcon, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

• Where did you find this problem? – Carl Schildkraut Feb 23 '17 at 22:06
• @CarlSchildkraut HMMT 2017 – Anonymous Feb 23 '17 at 22:15
• Harvard MIT Math Tournament? Is that ongoing? – Gerry Myerson Feb 23 '17 at 22:16
• @GerryMyerson It was last Saturday. – Carl Schildkraut Feb 23 '17 at 22:51

Let $\nu_p$ denote the $p$-adic order, i.e. $\nu_p(x)$ is the largest $i$ such that $p^i \mid x$.
Any prime that divides $a$ divides $b$, and any prime that divides $b$ divides $a$. If $2017 \nu_p(a) \ne \nu_p(b)$, then $\nu_p(a^{2017}+b) = \min(2017 \nu_p(a), \nu_p(b) \le \nu_p(b)$, while $\nu_p(ab) = \nu_p(a) + \nu_p(b)$, so it's impossible for $ab \mid a^{2017} + b$. Thus we must have $2017 \nu_p(a) = \nu_p(b)$. Since this is true for all $p$ dividing $a$ or $b$, $b = a^{2017}$. Now our condition is
$$a^{2018} \mid 2 a^{2017}$$
and so the only solutions are $a = b = 1$ and $a = 2$, $b= 2^{2017}$.