The proof was given as:
- Let $p\in\mathbb{N}^{+}$
- Let $a\in\mathbb{N}$
- Let $p$ be prime
- We know that $p\,\vert\,a^p-a$ by proposition of Chapter 1
- Therefore $a^p-a\equiv 0\pmod p$ by Definition 8.4.1
- Therefore $a^p\equiv a\pmod p$ by Proposition 8.4.7
Q.E.D.
But how do I get from 4 to 5 and 5 to 6?
Is 4 to 5 trying to say, since $p$ divides $(a^p - a)$ then the remainder is always 0? Then 5 - 6, how do I just move the $a$ over?
1 (original question)