# Odd Natural number and perfect squares

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

• In terms of $n$, what is this perfect square that should lie between $n^2$ and $2n^2$? – Andrés E. Caicedo Jul 20 '14 at 17:59
• 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? – James Jul 20 '14 at 18:00
• 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. – Thomas Andrews Jul 20 '14 at 18:04
• And if there were more than one, it would include the next two. – John Hughes Jul 20 '14 at 18:07

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.
• if $(n+1)^2>2n^2$ we get $0$ perfect squares between $n^2$ and $2n^2$ – Jorge Fernández Hidalgo Jul 20 '14 at 18:35