Integer to Power of Multiple of Order

Theorem
Let $$a$$ and $$n$$ be integers.

Let $$c \in \Z_+$$ be the order of $a$ modulo $n$.

Then $$a^k \equiv 1 \left({\bmod\, n}\right)$$ iff $$k$$ is a multiple of $$c$$.

Further, $$\phi \left({n}\right)$$ is a multiple of $$c$$, where $$\phi \left({n}\right)$$ is the Euler phi function of $$n$$.

Proof

 * Let $$k$$ be a multiple of $$c$$.

Then $$k = cr$$, say.

Then $$a^k = a^{cr} = \left({a^c}\right)^r \equiv 1^r \equiv 1 \left({\bmod\, n}\right)$$.


 * Let $$a^k \equiv 1 \left({\bmod\, n}\right)$$.

Then $$1 \equiv a^k \equiv a^{q c + r} \equiv \left({a^c}\right)^q a^r \equiv 1^q a^r \equiv a^r \left({\bmod\, n}\right)$$.

So $$r = 0$$ or else (from the Division Theorem) $$0 < r < c$$ and this contradicts $$c$$ being the smallest integer such that $$a^c \equiv 1 \left({\bmod\, n}\right)$$.

Hence $$k$$ is a multiple of $$c$$.

From Euler's Theorem, we have $$a^{\phi \left({n}\right)} \equiv 1 \left({\bmod\, n}\right)$$ and the above result can be applied directly.