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The problem below is from Cupillari's Nuts and Bolts of Proofs.

Prove the following statement:

Let $a$ and $b$ be two relatively prime numbers. If there exists an $m$ such that $(a/b)^m$ is an integer, then $b=1$.

My question is: Is the statement true?

I believe the statement is false because there exists an $m$ such that $(a/b)^m$ is an integer, and yet $b$ does not have to be $1$. For example, let $m=0$. In this case, $(a/b)^0=1$ is an integer as long as $b \neq 0$.

So I think the statement is false, but I am confused because the solution at the back of the book provides a proof that the statement is true.

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Presumably the problem meant that $m>0$ and $m$ is an integer. It's not true if $m<0$ either, since $(\frac{1}{2})^{-1}$ is an integer. – Thomas Andrews Nov 8 '11 at 19:28
(Actually, it's true for rational $m>0$, not just integer $m$.) – Thomas Andrews Nov 8 '11 at 19:31
@thomas Thank you! This gives me some confidence to fill in gaps in other incomplete problem statements I might encounter. – Andrew Liu Nov 8 '11 at 19:39
I think Cupillari should have said "positive integer". – Michael Hardy Nov 8 '11 at 20:21
up vote 3 down vote accepted

Your counterexample is valid. But the statement is true if $m$ is required to be a positive natural number or positive integer.

Alternatively, note it's not if true $m$ is required to be negative.

In my opinion, it seems like you were supposed to assume $m>0$.

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This statement is true under certain conditions. You must assume $b \neq 0$. If $(a,b)=1$, you can easily show that $(a^n, b^n) = 1$. This means that $(a/b)^n = a^n / b^n$ is never an integer unless there exists $n$ such that $b^n = \pm 1$, but that means $b = \pm 1$.

The case where $a = 0$ and $|b| > 1$ must be excluded because then $(a,b) > 1$.

Hope that helps,

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