I understand that any natural number congruent to 2 or 4 (mod6) is even and that 3 is the only prime congruent to 3(mod6). Thus, any prime, p > 3, must be congruent to either 1 or 5 (mod 6).
I just don't understand the reasoning behind why any natural number congruent to 2 or 4 (mod6) is even and that any natural number congruent to 3(mod6) is a multiple of 3.
Could someone explain why this is true?
All you have to do is eliminate the other cases, and use the definitions of modular congruence, divisibility, and primality. Recall:
$$x \equiv y \pmod z \iff z \mid x-y \iff \exists k \in \Bbb Z^+ \text{ such that } x-y = kz$$
(Or at least, we use the positive integers in this context usually.)
So, if $p \equiv 0 \pmod 6$, then $p = 6k$ for an integer $k$. That means $p$ isn't prime.
If $p \equiv 3 \pmod 6$, then $p = 6k+3$ for an integer $k$. But then $p=3(k+1)$, and thus is not prime (assuming $p > 3$).
If $p\equiv 2,4 \pmod 6$ then you have that $p=6k+2 = 2(3k+1)$ or $p = 6k+4 = 2(3k+2)$, respectively, for integers $k$, and thus no primality (assuming $p > 3$).
Thus, the only remaining cases are $1$ and $5$.
