# Solve $px + y^{p-1} = 2017$ [closed]

$p$ is a prime positivie number . $(x,y) \in \mathbb {N}^2$ . Solve $px + y^{p-1} = 2017$

## closed as off-topic by choco_addicted, Leucippus, Davide Giraudo, kingW3, NamasteJun 9 '17 at 13:29

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• A trivial solution would be $p = 2$, $y = 2017 - 2x$, and $x$ is any positive integer such that $y$ is also positive. Also, what have you tried? – Tob Ernack Jun 8 '17 at 23:55
• Another trivial solution is $p=2017$, $y=0$, and $x=1$ if you include zero in the natural numbers – Christian Woll Jun 8 '17 at 23:57
• Is this a question from an on-going contest? – Joel Reyes Noche Jun 9 '17 at 3:50

There are a lot of solutions. We'll use a few cases:

$y=0$: We have $xp=2017$ which has one solution: $x=1$ and $p=2017$

$y=1$: We have $xp=2016$ which has one solution: $x=1008$ and $p=2$

$y \ge 2$:

$\quad$Because $y^{p-1}\le 2017$, it follows that $$p-1 \le \text{log}_2(2017) \ \Rightarrow\ p \le11$$ When $p=2$, we have a trivial solution with $y=2017-2x$ and any $x \in[0,1008]$

A computer search reveals the remaining solutions: $(x,y,p)= (27, 44, 3),(56, 43, 3),(112, 41, 3),(139, 40, 3),(191, 38, 3),(216, 37, 3),(264, 35, 3),(287, 34, 3),(331, 32, 3),(352, 31, 3),(392, 29, 3),(411, 28, 3),(447, 26, 3),(464, 25, 3),(496, 23, 3),(511, 22, 3),(539, 20, 3),(552, 19, 3),(576, 17, 3),(587, 16, 3),(607, 14, 3),(616, 13, 3),(632, 11, 3),(639, 10, 3),(651, 8, 3),(656, 7, 3),(664, 5, 3),(667, 4, 3),(671, 2, 3),(672, 1, 3),(184, 3, 7),(279, 2, 7),(288, 1, 7)$

That makes $1043$ distinct solutions.

Hint (from Fermat's little theorem, if $y$ does not contain $p$): $$y^{p-1} \equiv^p 1 \Rightarrow 2017 \equiv^p 1 \Rightarrow p| 2016 \Rightarrow p = 2, 3, 7$$

I should have mentioned that if $p|y$, therefor $p|2017$. As 2017 is prime, $p=2017$.