Prove that $m = \sqrt {n +\sqrt {n +\sqrt {n + \cdots}}}$ is an integer if and only if $n$ is equal to twice a triangular number. $m = \sqrt {n +\sqrt {n +\sqrt {n + \cdots}}}$ 
Prove that $m$ is an integer if and only if $n$ is equal to twice a triangular number.
Supposing that $n$ is twice a triangular number. Then I have that $m = \sqrt {2t +m}$  for some traingular number $t$ thus $m^2-m-2t=0$ so, by the quadratic formula $m=\dfrac{1+\sqrt{8t+1}}{2}$
So I need to show that $8t +1$ is a square for any $t$. If so than it's square root is clearly an odd number so $1+\sqrt{8t+1}$ is clearly even and therefore divisible by $2$. So $m$ is an integer.
But I'm unsure how to show this.

We have 
\begin{eqnarray*}
m= \sqrt{ n+ \sqrt{n+ \sqrt{n+ \cdots}}} \\
\end{eqnarray*}
So $m= \sqrt{n+m}$. Square this and solve the quadratic
\begin{eqnarray*}
m= \frac{-1 + \sqrt{1+4n}}{2}
\end{eqnarray*}
So $1+4n$ will need to be a perfect square to deal with the square root; indeed a square of an odd value to deal with the division by $2$. So $1+4n =(2k+1)^2$.
