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While playing with numbers, I discovered a relationship which I would like to verify here.

$a^p\mod p=a \mod p$ when $p$ is prime

Proof:

If $c$ is prime number, then;

$a^c\mod{p}=\mod(\mod (b^c)+\mod(d^c))$ where $b+d=a,$

By splitting $b$ and $c$ in the same way, we arrive at:

$a^p \mod p=\mod(b+d)$

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Curious
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    Your proof needs work, but the result is correct, and is called Fermat's Little Theorem. As it is, your sentences have bad syntax, and have no meaning. – GFauxPas Dec 28 '14 at 15:23
  • We split $b$ as ($1^c+1^c+1^c..$){$b$ times} and $d$ as ($1^c+1^c+...$){$d$ times) – Curious Dec 28 '14 at 15:28
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    Look at your sentences carefully. They don't make sense. – GFauxPas Dec 28 '14 at 15:30
  • I think that your idea of proof is the (well known, sorry) proof by induction: we have $1^p=1\mod{p}$, and using the fact that $\binom{p}{k}=0\mod{p}$ for $1\leq k\leq p-1$, we have $(a+1)^p=a^p+1^p \mod {p}$, (hence equal $a+1 \mod{p}$) by the binomial theorem. – Kelenner Dec 28 '14 at 15:45

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