Suppose a, b and n are positive integers. Prove that (a^n) | (b^n) if and only if a | b.

Question

Suppose $a, b$ and $n$ are positive integers. Prove that $a^n\mid b^n$ if and only if $a \mid b$.

I have:

$$a^n\mid b^n$$

$$\implies b^n = a^n \cdot k$$

$$\implies \sqrt[n]{b^n}=\sqrt[n]{a^n}\cdot k$$

$$\implies a=b\cdot k$$

$$\implies a\mid b$$

Is it really this simple?

You need to take the nth root of all of the right-hand side, and there's no way of guaranteeing that $\sqrt[n] k$ is an integer: $$\sqrt[n]{b^n} = \sqrt[n]{a^nk} \iff b = a\sqrt[n] k$$ — amWhy, Nov 18 '13 at 23:32

score 0 · Accepted Answer · answered Nov 18 '13 at 23:52

Factor $a$ and $b$ to prime factors: $$ a = 2^{a_1} 3^{a_2} 5^{a_3} 7^{a_4} \dots $$

$$ b = 2^{b_1} 3^{b_2} 5^{b_3} 7^{b_4} \dots $$

Now

$$ a^n = 2^{na_1} 3^{na_2} 5^{na_3} 7^{na_4} \dots $$

$$ b^n = 2^{nb_1} 3^{nb_2} 5^{nb_3} 7^{nb_4} \dots $$

We know that

$$ a | b \Leftrightarrow \forall i \in \mathbb N : a_i \le b_i $$

And finally

$$ \forall x, y, n \in \mathbb N : x \le y \Leftrightarrow x n \le y n $$

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