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.

Question

$m = \sqrt {n +\sqrt {n +\sqrt {n + \cdots}}}$

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.

Answer 1

HINT: $$8t + 1 = 8 \cdot \frac{s(s-1)}{2} + 1 = 4s^2 - 4s + 1 = (2s - 1)^2$$

Answer 2

If $m = \sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}$, then $m = \sqrt{n+m}$ so $n = m^2 - m = 2\cdot\dfrac12(m-1)(m)$ is twice a triangular number.

Answer 3

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 $1+4n =(2k+1)^2$.