# 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

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• 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 $a = 2$, $b= 2^{2017}$.