How to prove that $\frac{1}{n+1} = {\sqrt{n\over n+1}}\implies n = \Phi$?

Question

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.

Answer 1

HINT:

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

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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