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By brute force I'm noticing that for all $a ∈ _7$, we have $a^{19}\bmod 7 \equiv a $. Now this looks very much like Fermat's Little Theorem, but this theorem should hold if the primes are the same, whereas 19 and 7 are clearly not. Is this just a coincidence?

Bill Dubuque
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Jos van Nieuwman
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  • It is FLT. If $7|a$ then $a^{19}\equiv 0\equiv a \pmod 7$ and if $7\not \mid a$ the $a^{6} \equiv 1\pmod 7$ so $a^{18}\equiv 1\pmod 7$ and $a^{19}\equiv a \pmod 7$. – fleablood Apr 18 '20 at 20:36

2 Answers2

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Since $7$ is prime, $\varphi(7)=6$, so $$a^6=1\pmod{7}$$ by Euler's Theorem, where $\varphi $ is Euler's totient function.

Shaun
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    This doesn't cover the case $a=0$, although that's almost trivial to remedy. Also, note that Fermat's little theorem is enough (not that I'm against Euler's theorem, it's just that FLT was mentioned in the OP, so it could be nice to keep using that). – Arthur Apr 18 '20 at 19:49
  • Thank you for making that explicit, @Arthur. – Shaun Apr 18 '20 at 19:50
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By Fermat's Little Theorem:

$a^{6}\equiv 1\pmod{7}$

Thus:

$(a^{6})^{3}\equiv 1^{3}\pmod{7}$

$a^{18}\equiv 1\pmod{7}$

Multiplying by $a$ on both dies, we have:

$\boxed{a^{19}\equiv a\pmod{7}}$

The above argument works only for $a$ relatively prime to $7$. For $a$ not relatively prime to $7$, this means $a\equiv 0\pmod{7}$ because $7$ is prime. Then, it can be demonstrated that

$0^{19}\equiv 0\pmod{7}$

Joshua Wang
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