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, Shailesh Jun 3 '17 at 0:41

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  • 1
    $\begingroup$ Where did you find this problem? $\endgroup$ – Carl Schildkraut Feb 23 '17 at 22:06
  • $\begingroup$ @CarlSchildkraut HMMT 2017 $\endgroup$ – Anonymous Feb 23 '17 at 22:15
  • $\begingroup$ Harvard MIT Math Tournament? Is that ongoing? $\endgroup$ – Gerry Myerson Feb 23 '17 at 22:16
  • $\begingroup$ @GerryMyerson It was last Saturday. $\endgroup$ – 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 $a = 2$, $b= 2^{2017}$.


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