Divisor of Integer/Examples/63 divides 8^2n - 1/Proof 1

Proof
Proof by induction:

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $63 \divides 8^{2 n} - 1$

$\map P 0$ is the case:

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:
 * $63 \divides 8^{2 k} - 1$

from which it is to be shown that:
 * $63 \divides 8^{2 \paren {k + 1} } - 1$

Induction Step
This is the induction step:

From the induction hypothesis we have that:
 * $63 \divides 8^{2 k} - 1$

Hence by definition of divisibility, we have:
 * $\exists r \in \Z: 8^{2 k} - 1 = 63 r$

and so:
 * $(1): \quad \exists r \in \Z: 8^{2 k} = 63 r + 1$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 0}: 63 \divides 8^{2 n} - 1$