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Conjecture:
$n^{2p-1}\equiv n\pmod {2p}$ for all $n\in \mathbb N$ and all odd primes $p$.

I started investigate the least $x_n$ such that $p_n^{p_{n+1}}\equiv x_n\pmod{p_{n+2}}$ and ended up with this for me unknown conjecture. I've tested it for a lot of random values of $n$ and $p$.

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closed as off-topic by user21820, Xander Henderson, José Carlos Santos, Lord Shark the Unknown, Jose Arnaldo Bebita-Dris Aug 3 '18 at 9:59

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    $\begingroup$ What is $x$ ? What is the conjecture ? $\endgroup$ – Donald Splutterwit Jun 27 '18 at 19:13
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    $\begingroup$ This follows from Little Fermat. Sketch: if $\gcd (n,p)=1$ then $n^{p-1}\equiv 1 \pmod p\implies n^{2p-2}\equiv 1 \pmod p\implies n^{2p-1}\equiv n\pmod p$. (it's obvious that the two sides have the same parity $\pmod 2$). $\endgroup$ – lulu Jun 27 '18 at 19:13
  • $\begingroup$ @donald: well, I investigated the least x... $\endgroup$ – Lehs Jun 27 '18 at 19:18
  • $\begingroup$ @lulu: Thanks for the explanation! $\endgroup$ – Lehs Jun 27 '18 at 19:22
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Surely $n^{2p-1}-n$ is always even then it remains to show that $p$ divides $n^{2p-1}-n$. If $p\mid n$ it's obvious and if $p\nmid n$ it is true according to Fermat's little theorem https://en.wikipedia.org/wiki/Fermat%27s_little_theorem

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    $\begingroup$ Tip: you can use \mid and \nmid for more nicely looking (non)divisibility relation: $p\mid n,p\nmid n$. $\endgroup$ – Wojowu Aug 3 '18 at 9:03
  • $\begingroup$ Thanks. Nice tip! $\endgroup$ – Mostafa Ayaz Aug 3 '18 at 9:04

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