Euler's Theorem

Theorem
Let $$a, m \in \mathbb{Z}: a \perp m$$.

Let $$\phi \left({m}\right)$$ be the Euler Phi Function of $$m$$.

Then $$a^{\phi \left({m}\right)} \equiv 1 \left({\bmod\, m}\right)$$.

Proof
Since $$a \perp m$$, the congruence class modulo $m$ of $a$ belongs to the Multiplicative Group of Integers Modulo m $$\left({\mathbb{Z}'_m, \times}\right)$$.

Let $$ k = \left|{\left[\!\left[{a}\right]\!\right]_m}\right|$$ where $$\left[\!\left[{a}\right]\!\right]_m \in \mathbb{Z}'_m$$.

By Order of Element Divides Order of Finite Group, $$k \backslash \left|{\mathbb{Z}'_m}\right|$$.

Now $$\left|{\mathbb{Z}'_m}\right| = \phi \left({m}\right)$$, from the definition of the Euler Phi Function.

Thus:

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