Suppose we have an element $a \in \mathrm{GF}(p)$ where $p$ is Prime. Is it true, that we always have $a^p = a \mod p$? If so, how would one go about proving this for all $p$?
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Do you already know Fermat's little theorem or Lagrange's theorem for groups? Please give further context. This will end up being a duplicate question in any case. – Bill Dubuque May 08 '22 at 11:29
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I didn't until now, but I guess Fermats little theorem comes in handy – Al Sneed May 08 '22 at 11:32
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Yes it is equivalent to Fermat's little theorem (see here for how to show equivalent the two common forms of little Fermat). – Bill Dubuque May 08 '22 at 11:53
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More generally in any finite field $\mathbb{F}_q$ (here $q=p^f$ with $p$ prime) one has $ a^q=a\quad\text{for all $a\in\mathbb{F}_q$.} $ Indeed the multiplicative group $\mathbb{F}_q$ has $q-1$ elements so that if $a\neq0$ a straightforward application of Lagrange's theorem for finite groups gives $a^{q-1}=1$ from which the claim follows at once.
Finally, $0^q=0$ is trivial.
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Andrea Mori
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1Please strive not to post more (dupe) answers to dupes of FAQs/PSQs cf. recent site policy announcement here. – Bill Dubuque May 08 '22 at 11:57