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

## Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Then:

$63 \divides 8^{2 n} - 1$

where $\divides$ denotes divisibility.

## 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:

 $\ds 8^{2 \times 0} - 1$ $=$ $\ds 8^0 - 1$ $\ds$ $=$ $\ds 1 - 1$ Zeroth Power of Real Number equals One $\ds$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds 63$ $\divides$ $\ds 8^{2 \times 0} - 1$ Integer Divides Zero

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

### Basis for the Induction

$\map P 1$ is the case:

 $\ds 8^{2 \times 1} - 1$ $=$ $\ds 8^2 - 1$ $\ds$ $=$ $\ds 64 - 1$ $\ds$ $=$ $\ds 63$ $\ds \leadsto \ \$ $\ds 63$ $\divides$ $\ds 8^{2 \times 1} - 1$ Integer Divides Itself

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:

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

 $\ds 8^{2 \paren {k + 1} } - 1$ $=$ $\ds 8^2 \times 8^{2 k} - 1$ $\ds \leadsto \ \$ $\ds \exists r \in \Z: \,$ $\ds 8^{2 \paren {k + 1} } - 1$ $=$ $\ds 64 \times \paren {63 r + 1} - 1$ from $(1)$ $\ds$ $=$ $\ds 64 \times 63 r + 63$ algebra $\ds$ $=$ $\ds 63 \paren {64 r + 1}$ algebra $\ds \leadsto \ \$ $\ds 63$ $\divides$ $\ds 8^{2 \paren {k + 1} } - 1$ Definition of Divisor of Integer

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

Therefore:

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

$\blacksquare$