If $a^6\mid b^6$, then does $a\mid b$?

Asked Jul 17 '19 at 23:36

Active Jul 17 '19 at 23:49

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If $a^6 \mid b^6$, where $a,b\in \mathbb{Z^{+}}$, then does $a\mid b$?

I was thinking of using the Fundamental Theorem of Arithmetic to show that $$a^6=\prod_{i}p_{i}^{6k_i}\quad\text{and}\quad b^6=\prod_{i}p_{i}^{6q_i}$$ And, since $6k_i\leq 6q_i$, $\forall i$, $k_i \leq q_i$ which implies $a\mid b$.

Is this correct?

edited Jul 17 '19 at 23:49

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asked Jul 17 '19 at 23:36

user100000001

You are right! I think it's the easiest way to see that. – azif00 Jul 17 '19 at 23:44
Yes, this is a correct way to solve it. – John Omielan Jul 17 '19 at 23:44
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Here is another cute way to solve it. Let $k=\frac{b^6}{a^6} \in \mathbb Z$. Then, $\frac{b}{a}$ is a rational root of $X^6-k \in \mathbb Z[X]$, and hence an integer root. – N. S. Jul 17 '19 at 23:53

If $a^6\mid b^6$, then does $a\mid b$?

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