# 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!

• Presumably, you don't define $\mathbb N$ as including $0$... – Thomas Andrews Oct 6 '15 at 2:30

## 3 Answers

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 Oct 6 '15 at 2:25
• Oh wow didn't see it was that simple, thanks so much!! – Twis7ed Oct 6 '15 at 2:26
• Great approach. Thanks – Shailesh Oct 6 '15 at 3:44

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...

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