For which values of m?

Question

I've been wondering about this for a while:

for which values of $m$ is $a^k\equiv a\pmod m$ for all $0\le a<m$. I know that $m=6$ works.

I meant $k$ is odd (for the $m=6$ case). Sorry about that.

See also http://math.stackexchange.com/questions/164524/largest-modulus-for-fermat-type-polynomial. — lhf, Dec 12 '13 at 13:32

score 1 · Answer 1 · edited Dec 09 '13 at 19:32

Assuming that $k$ is fixed and $k$ is not $1$. This reduces your question to one of reciprocity.

You are asking whether $(a/m)_{k-1} =1$ or not. If you examine a primitive character mod $m$ you need to know its order, and its order would divide $\phi(m)$, but there is not character $\chi$ such that $\chi(a)=1$ for all $a (\mod m)$ (unless it is the primitive character) hence there are no such values of m, unless m=1.

score 1 · Answer 2 · answered Dec 09 '13 at 20:20

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I am sorry about my previous answer, there was an error, in there.

The values of such m are those of the form $m=2p$ where $p$ is some odd prime. this is because $\varphi(m)=\varphi(p)$ and so the value of $k$ in your problem will always be $p$ hence you would have $a^p\equiv a(\mod 2p)$

answered Dec 09 '13 at 20:20

user114698

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Occasionally it is helpful to post a second Answer, e.g. to show an alternative approach to (or interpretation of) a Question. However if what is desired is to post a correction to a previous Answer, this should be done simply by editing that Answer. – hardmath Dec 10 '13 at 04:30

For which values of m?

2 Answers2