I noticed that for any integer greater than 4 there exists at least one perfect square in the open interval ((adsbygoogle = window.adsbygoogle || []).push({}); n,2n). I think I have proved the statement, but as I am not a professional, I'd like someone to review my proof.

By induction we see that it is true for [itex]n=5[/itex] because [itex]5<9<10[/itex].

Now suppose that it is true forn, and let's prove it forn+ 1.

From the inductive hypothesis we know that there is anxsuch that [itex]n<x^2<n+1[/itex]. This square should work forn+ 1 too. Like in the case of [itex]n=5[/itex] the square 9 works for 6, 7 and 8 too. But not for [itex]n=9[/itex] or greater. Here we need the next square 16.

So if [itex]x^2\ge n+1[/itex] we must show that [itex](x+1)^2[/itex] is the square we need. In other words we have to verify that

[tex]n+1<(x+1)^2<2(n+1)[/tex]

Notice that for positive n, if [itex]x^2\le n+1[/itex] then [itex]x\le \sqrt{n+1}[/itex]. As [itex]n<x^2[/itex] it follows that

[tex]n+1<x^2+1<x^2+2x+1=(x+1)^2[/tex]

We have to show that [itex](x+1)^2<2(n+1)[/itex]

[tex](x+1)^2=x^2+2x+1<\left(n+1\right)+2\sqrt{n+1}+1[/tex]

[tex]\left(n+1\right)+2\sqrt{n+1}+1 < 2(n+1)[/tex]

[tex]n^2-4n-4>0\rightarrow n>2\sqrt 2 + 2\rightarrow n\ge 5[/tex]

So we proved that for any [itex]n\ge 5[/itex] there is at least onexsuch that [itex]n< x^2 < 2n[/itex].

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# Square numbers between n and 2n -- Check my proof please

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