# Powers of Coprime Numbers are Coprime

## Theorem

Let $a, b$ be coprime integers:

$a \perp b$

Then:

$\forall n \in \N_{>0}: a^n \perp b^n$

In the words of Euclid:

If two numbers be prime to one another, and each by multiplying itself make a certain number, the products will be prime to one another; and if the original numbers by multiplying the products make certain numbers, the latter will also be prime to one another [and this is always the case with the extremes].

## Proof

Proof by induction:

Let $a \perp b$.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$a^n \perp b^n$

$\map P 1$ is true, as this just says:

$a \perp b$

### Basis for the Induction

$a^2 \perp b$
$a^2 \perp b^2$

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$a^k \perp b^k$

Then we need to show:

$a^{k + 1} \perp b^{k + 1}$

### Induction Step

This is our induction step:

 $\ds a^k$ $\perp$ $\ds b^k$ $\ds \leadsto \ \$ $\ds a^k \times a$ $\perp$ $\ds b^k$ Proposition $24$ of Book $\text{VII}$: Integer Coprime to all Factors is Coprime to Whole $\ds \leadsto \ \$ $\ds a^{k + 1}$ $\perp$ $\ds b^k$ $\ds \leadsto \ \$ $\ds a^{k + 1}$ $\perp$ $\ds b^k \times b$ Proposition $24$ of Book $\text{VII}$: Integer Coprime to all Factors is Coprime to Whole $\ds \leadsto \ \$ $\ds a^{k + 1}$ $\perp$ $\ds b^{k + 1}$

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

Therefore:

$\forall n \in \N_{>0}: a^n \perp b^n$

$\blacksquare$

## Historical Note

This proof is Proposition $27$ of Book $\text{VII}$ of Euclid's The Elements.
Euclid's proof does not use the full induction process, which is a later mathematical innovation.