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

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Theorem

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

Then:

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

where $\divides$ denotes divisibility.


Proof 1

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:

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

$\blacksquare$


Proof 2

From Integer Less One divides Power Less One, we have that:

$\forall m, n \in \Z: \paren {m - 1} \divides \paren {m^n - 1}$

This result is the special case where $m = 8^2$.

$\blacksquare$

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