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Can someone help me prove/disprove this?

I wrote $n^6$ as $l^2$, but I don't know how to convert $n^{50}$ into that format because $^{50}$ is too large.

$n$ ∈ ℤ

user272513
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  • What kind of number is $n$? – ajotatxe Sep 20 '15 at 16:38
  • Sorry I forgot to add that, $n$ ∈ ℤ – user272513 Sep 20 '15 at 16:40
  • Typesetting note, just use { } in order to have longer exponents. n^{reallylong} gives $n^{reallylong}$ as opposed to n^reallylong which gives $n^reallylong$ – JMoravitz Sep 20 '15 at 16:41
  • But $n^6=(n^3)^2$ is always a perfect square... – Intelligenti pauca Sep 20 '15 at 16:43
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    About the question, if $n$ is indeed an integer, then it doesn't matter that $n^6$ is a perfect square since $n^{50}$ is always going to be a perfect square anyways. The more interesting question is if $n$ is not necessarily an integer but it turns out that $n^6$ is a perfect square, for example $n=\sqrt[3]{2}$. In this case, what happens? $(\sqrt[3]{2})^6=4$ and $(\sqrt[3]{2})^{50}=?$ – JMoravitz Sep 20 '15 at 16:44
  • @JMoravitz: One could make it fancier by asking, given that $x^{50}$ is an integer, whether it is a perfect square. – André Nicolas Sep 20 '15 at 17:02
  • @AndréNicolas Hmm, interesting. Mightn't that make it obviously true when $n$ is an integer and vacuously true when $n$ is not? $n^6$ being a perfect square implies that $n=p_1^{\frac{\alpha_1}{3}}p_2^{\frac{\alpha_2}{3}}\dots$ and is an integer if and only if $3\mid \alpha_i$ for each $i$. When $n$ is not an integer, that means that there is some $\alpha_i$ that is not a multiple of three, which after multiplication by $50$ will again not be a multiple of $3$, implying that $n^{50}$ would not be an integer. – JMoravitz Sep 20 '15 at 17:14
  • Sure, the point is one has to work. Reflex action on my part, trying to make up a "new" test/homework question. – André Nicolas Sep 20 '15 at 17:17

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Both $n^6$ and $n^{50}$ are perfect squares, since $n^6=(n^3)^2$ and $n^{50}=(n^{25})^2$. What's the problem?