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How can i proof that $41$ does not divide $a^5 - 2$ for any $a \in \Bbb Z$?

I think i must show that $a^5 - 2$ is multiple of 41.

So if i do $(a^5 - 2) / 41$ i must get 0 rest?

I'm stuck sorry.

Thank you.

Mudasty
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    Do you know Fermat’s little theorem? – J. W. Tanner May 05 '21 at 22:39
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    Welcome to stackexchange. This is not a "do this for me" site. If you [edit] the question to show us what you tried and where you are stuck we may be able to help. – Ethan Bolker May 05 '21 at 22:39
  • If $p$ is a prime and $a$ is any integer not divisible by p, then $a^{p-1} − 1$ is divisible by $p$. My question is how to prof the specific $a^5-2$. thankyou – Mudasty May 05 '21 at 22:47
  • What have you tried? EG Did you brute force check $ a =0, 1, 2, \ldots 40$? – Calvin Lin May 05 '21 at 22:50
  • isn't a multiple * 2 would have to be a fifth power remainder of division by 41 for it be a multiple, but you can do no more than 20 cases and figure if it is if you know enough. – Roddy MacPhee May 06 '21 at 01:46

1 Answers1

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If we were to have $a^5\equiv2\bmod41$,

then we would have $1\equiv a^{40}\equiv2^8=256=246+10\equiv10\bmod41$ -- a contradiction.

J. W. Tanner
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