Corollary of Fermat's Little Theorem to $a^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1}$

Question

I'm not sure what to call this proposition in relation to Fermat's Little Theorem (it is not the converse, though it seems related), but I am interested to know if the following holds:

$$ a^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1} $$ for $p$ prime and $a \neq 1$.

If this does hold, what is the proof? If not, are there any conditions on $x$ required in order to have $a^x \equiv 1 \pmod{p}$?

We need this to be true for all $a$ (or at least for carefully chosen $a$) to get the conclusion. Otherwise, not only is $a=1$ a counterexample, but so are more subtle cases like $13^4 \equiv 1 \pmod {17}$. — Misha Lavrov, Mar 31 '17 at 20:20
Sorry, I meant to have $a \neq 1$. I will edit the question. — Alex, Mar 31 '17 at 20:21

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

For each prime $p$, there exists a primitive root $r$ which has the property you want: $$r^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1}.$$ For example, if we work modulo $7$, $3$ is a primitive root, because \begin{align} 3^0 &\equiv 1 \pmod 7 \\ 3^1 &\equiv 3 \pmod 7 \\ 3^2 &\equiv 2 \pmod 7 \\ 3^3 &\equiv 6 \pmod 7 \\ 3^4 &\equiv 4 \pmod 7 \\ 3^5 &\equiv 5 \pmod 7 \\ 3^6 &\equiv 1 \pmod 7 \\ \end{align} so the powers of $3$ have a period of exactly $6$. It follows that $2$ will not work mod $7$, because $$2^3 \equiv (3^2)^3 \equiv 3^6 \equiv 1 \pmod 7.$$

This MSE post gives a short proof of the existence of primitive roots modulo any prime $p$.

Corollary of Fermat's Little Theorem to $a^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1}$

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