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Find all odd natural numbers $n$ for which there is a unique perfect square strictly between $n^2$ and $2n^2$. I considered some numerical examples $3$ is and odd number and between $3^2$ and $2(3)^2$. i.e., between 9 and 18 the unique square is 16. It is true. How should I begin the proof

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    $\begingroup$ In terms of $n$, what is this perfect square that should lie between $n^2$ and $2n^2$? $\endgroup$ – Andrés E. Caicedo Jul 20 '14 at 17:59
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    $\begingroup$ You are trying to find unique perfect squares between $n^2$ and $2n^2 = (\sqrt{2}n)^2$. So it is the same as finding a unique integer between $n$ and $\sqrt{2}n$. Does that help? $\endgroup$ – James Jul 20 '14 at 18:00
  • $\begingroup$ What is the next perfect square after $n^2$? If there is any squares between $n^2$ and $2n^2$, it would have include the next highest square. $\endgroup$ – Thomas Andrews Jul 20 '14 at 18:04
  • $\begingroup$ And if there were more than one, it would include the next two. $\endgroup$ – John Hughes Jul 20 '14 at 18:07
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basically this question asks for what odd value of $n$ do we have $2n^2\leq(n+2)^2=n^2+4n+4$

So we first solve the inequality: $2n^2\leq n^2+4n+4 \iff n^2<4n+4\iff n\leq4+\frac{4}{n}\implies n\leq4$

so the only values can be $1$ or $3$. Check there is no perfect square between 1 and 2, and check $3$ works.

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  • $\begingroup$ if $(n+1)^2>2n^2$ we get $0$ perfect squares between $n^2$ and $2n^2$ $\endgroup$ – Jorge Fernández Hidalgo Jul 20 '14 at 18:35
  • $\begingroup$ Oh, he wants a unique perfect square. I can't read. $\endgroup$ – Thomas Andrews Jul 20 '14 at 18:39
  • $\begingroup$ ok, thank you, I changed it $\endgroup$ – Jorge Fernández Hidalgo Jul 20 '14 at 18:43

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