Natural Number has Same Prime Factors as Integer Power

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Theorem

Let $x$ be a natural number such that $x > 1$.

Let $n \ge 1$ be a (strictly) positive integer.

The $n$th power of $x$ has the same prime factors as $x$.

Proof

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Let $p$ a prime number such that $p$ divides $x^n$.

This is possible because $x > 1$, so $x^n > 1$, hence $x^n$ has prime divisors due to Fundamental Theorem of Arithmetic.

To prove the statement, we need to show $p$ divides $x$.

We will prove this statement by the Principle of Mathematical Induction on $n$.

Basis of the Induction

We have $n = 1$

Clearly, since $p$ divides $x^1 = x$, then $p$ divides $x$.

$\Box$

Inductive Step

Suppose that for a given $n$, if $p$ divides $x^n$ then $p$ divides $x$.

Then, if $p$ divides $x^{n+1}$, by definition of prime number, either $p$ divides $x^n$ or $p$ divides $x$.

If $p$ divides $x^n$, we get from induction hypothesis that $p$ divides $x$.

The other case trivially leads to our conclusion.

$\Box$

Hence the result, by Principle of Mathematical Induction.

$\blacksquare$

Conversely, let $p$ a prime number such that it divides $x$. Then, $p$ divides $x * x^{n-1} = x^n$, as required.

$\blacksquare$