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# Prove that for all $n\in\mathbb{N}$, $\sqrt{n(n+1)}$ is not an integer.

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!

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 $n\in\Bbb{N}$

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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# Prove that for all $n\in\mathbb{N}$, $\sqrt{n(n+1)}$ is not an integer.

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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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 $n\in\Bbb{N}$

Neat approach. I was going to suggest appealing to the Fundamental Theorem of Arithmetic but this is even better. - Cameron Williams yesterday

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