Suppose $n$ is an odd integer and $k \in \mathbb{N}$ with $k \geq 2$. How can I show the following statement?
$$ \sqrt{n}^k \ \text{is an integer}\ \Longrightarrow n \mid\sqrt{n}^k $$
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Please share your thoughts so far :) – Shaun Mar 07 '15 at 12:58
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It is obviously true when $k$ is even. The case for odd $k$ is only slightly more involved. Neither case seems to need the assumption that $n\ge 3$, they are equally true for $n=0,1,2$. – hmakholm left over Monica Mar 07 '15 at 13:10
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Let $p$ a prime dividing $n$. Then $p$ has to be odd. Let $a$ the biggest integer such that $p^a$ divides $n$. Then, we have equivalently to show that $$\frac{ak}{2} \in \mathbf{Z} \implies a \le \frac{ak}{2}.$$ Well, the last inequality holds if and only if $k\ge 2$, which is true by assumption.
Paolo Leonetti
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It is obvious for even $k$.
For odd $k$, we see that the powers of all prime factors of $n$ must be even, i.e, $n$ is a perfect square.
let $n = t^2$.
Therefore the statement boils down to $t^2\mid t^k$ , which is true for all $k\geq 2$.
Bayern ROX
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Obviously $\sqrt{n}^k=n\cdot\sqrt{n^{k-2}}$.
From the hypotheses, $\sqrt{n^{k-2}}$ is an integer. We know that the square root of a nonnegative integer is either an integer or an irrational number. (This is a well-known adaptation of the proof that the square root of two is irrational.) If $\sqrt{n}^k$ is an integer then our equation shows that $\sqrt{n^{k-2}}$ cannot be irrational, so it must be an integer. Therefore $n$ divides $n\cdot\sqrt{n^{k-2}}$ and thus $n$ divides $\sqrt{n}^k$.
Note that this proof does not use the hypothesis that $n$ is odd. It also does not use the hypothesis in the original version of the OP that $n\ge 3$.
Rory Daulton
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$ x = n^{k/2}/n \in \Bbb Q\,\Rightarrow\, x^2 = n^{k-2}\in \Bbb Z\,\Rightarrow\, x\in \Bbb Z\,$ by the Rational Root Test. $ $ Thus $\ n\mid n^{k/2}$
Bill Dubuque
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