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I have to find out under what conditions for $x$ the expression $\sqrt{x}/4$ is in $\mathbb Z$. So my idea was following:

$$\begin{align}\sqrt{x}/4 \in \mathbb Z \\ \iff \sqrt{x} \in 4\mathbb Z \\ \iff \sqrt{x} = 4n \text{ for some } n \in \mathbb Z \end{align}\\ \iff x = 16n^2 \text{ for some }n \in \mathbb Z$$

I am not sure about the last step, is that one true?

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    $\begingroup$ The condition is correct. $\endgroup$ – André Nicolas Jun 1 '15 at 15:36
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    $\begingroup$ @Martigan No, it says 'for some $n\in\Bbb Z$'. The 'some $n\in\Bbb Z$' there would be $-n$. We have $\sqrt{x}=4(\pm n)\iff x=16n^2$, and the equivalence is correct. $\endgroup$ – user26486 Jun 1 '15 at 15:39
  • $\begingroup$ Thanks AndréNicolas and user26486! $\endgroup$ – flawr Jun 1 '15 at 15:42
  • $\begingroup$ @user26486 No, I was talking about the "equivalence". You can't write, in a general sense, that $x=\alpha^2$ is equivalent to $\sqrt x=\alpha$... And in fact that is what you wrote by stating the $\pm$ sign in your comment... Your equivalence is correct, not the OP's. The fact that it is written for some $n \in \mathbb{Z}$ is not enough... $\endgroup$ – Martigan Jun 1 '15 at 15:51
  • $\begingroup$ @Martigan Again, it says 'for some $n\in\Bbb Z$'. $\pm n$ is an integer, whether it is $n$ or $-n$. $\endgroup$ – user26486 Jun 1 '15 at 15:56
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Your approach is correct. Here's the proof of the equivalence: $x$ is in the form $16n^2$ for some $n \in \mathbb Z$ $\iff$ $\sqrt x/4 \in \mathbb Z$:

  • Let $x$ be in form $16n^2$ for some $n$ from $\mathbb Z$. Then, square root of $x$ is either $4n$ or $-4n$ and $x/4$ is either $n$ or $-n$, which is, in any case, in $\mathbb Z$.

  • Now, let's take an $x$ such that $\sqrt x / 4$ is in $\mathbb Z$. We must find an $n \in \mathbb Z$ such that the $x$ is in the form $16n^2$. If we take $n=\sqrt x/4$, then $16n^2 = 16(\sqrt x/4)^2 = 16(x/16) = x$

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  • $\begingroup$ Ah now I did not see the first point but now it is obvious, thanks! $\endgroup$ – flawr Jun 1 '15 at 16:18

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