How to prove that $\frac{1}{n+1} = {\sqrt{n\over n+1}}\implies n = \Phi$? Consider:
$$\frac{1}{n+1} = {\sqrt{n\over n+1}}$$
How could one prove that $n$ is of such form that:
$$\frac{1}{n+1} = {\sqrt{n\over n+1}}\implies n = {\sqrt{5\,}-1 \over 2} \implies n = \Phi$$
where $\Phi$ denotes the golden ratio?
Edit:
I noticed that:
$$\frac{1}{n+1} = {\sqrt{n\over n+1}} \implies \left ( \frac{1}{n+1}\right ) ^2 = \left ({\sqrt{n\over n+1}}\right)^2\implies$$
$${1\over(n+1)^2} =  {n\over(n+1)}\implies n = {1\over n+1} \implies$$
$${1\over{1+{1\over1+n}}} \approx \Phi$$
as $$\Phi = {1\over{1+{1\over{1+{1\over\dots}}}}}$$
which should be proof enough.

HINT:  
Just square both sides to get $$\frac1{(n+1)^2}=\frac n{n+1}\implies n^2+n-1=0$$ assuming $n+1\ne0$

Alternatively, $\sqrt{n+1}=(n+1)\sqrt n\implies 1=\sqrt{n(n+1)}$ assuming $n+1\ne0$
Now, square both sides to get $n^2+n=1$
So, $n=\frac{-1\pm\sqrt5}2$
As $\sqrt{\frac n{n+1}}$ is considered to be $\ge 0,\frac1{n+1}\ge 0\implies n\ge -1$
Now, as $-\frac{1+\sqrt5}2<-1,$ it is an extraneous root
