Powers of 3 Modulo 8/Proof 1

Proof
Let the statement be rewritten as:

For all $r \in \Z_{\ge 0}$:
 * $3^r \equiv \begin {cases} 1 \pmod 8 & : r = 2 n \\ 3 \pmod 8 & : r = 2 n + 1 \end {cases}$

where $n \in \Z_{\ge 0}$.

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $3^{2 n} \equiv 1 \pmod 8$

$\map P 0$ is the case:
 * $3^{2 \times 0} = 3^0 = 1 \equiv 1 \pmod 8$

Thus $\map P 0$ is seen to hold.

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $3^{2 k} \equiv 1 \pmod 8$

from which it is to be shown that:
 * $3^{2 \paren {k + 1} } \equiv 1 \pmod 8$

Induction Step
This is the induction step:

We have:

So $\map P k \implies \map P {k + 1}$ and then it follows by the Principle of Mathematical Induction that:


 * $\forall n \in \Z_{\ge 0}: 3^{2 n} \equiv 1 \pmod 8$

Then we have:

and the result follows.