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I'm sure this is a very simple proof but I can't seem to get it right. I tried to do it by induction but get stuck trying to show that $\sqrt{(k+1)(k+2)}$ is not an integer and also cannot seem to do it through other methods. Anyone have any ideas? Thanks so much in advance!

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    $\begingroup$ Presumably, you don't define $\mathbb N$ as including $0$... $\endgroup$ – Thomas Andrews Oct 6 '15 at 2:30
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Despite being about $n\in\Bbb{N}$, induction doesn't work too great here.

Instead, how about trying to identify the nearest integers... in particular, you should be able to show $$n<\sqrt{n(n+1)}<n+1$$ for $n\in\Bbb{N}$

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  • $\begingroup$ Neat approach. I was going to suggest appealing to the Fundamental Theorem of Arithmetic but this is even better. $\endgroup$ – Cameron Williams Oct 6 '15 at 2:25
  • $\begingroup$ Oh wow didn't see it was that simple, thanks so much!! $\endgroup$ – Twis7ed Oct 6 '15 at 2:26
  • $\begingroup$ Great approach. Thanks $\endgroup$ – Shailesh Oct 6 '15 at 3:44
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Related approach: If $n(n+1)$ is a square, then $4(n^2+n)=4n^2+4n$ is a square. But so is $4n^2+4n+1=(2n+1)^2$. The difference of two square is $1$ when...

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$(n+1)^2 > n(n+1) > n^2$ taking square roots $n+1 > \sqrt{n(n+1)} > n$, and there is no integer between $n$ and $n+1$.

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