<s>[INST] Prove $1+\frac{1}{\sqrt{\phi^{4n+1}{F_{2n}F_{2n+1}}}+\phi^{2n+1}F_{2n}}=\sqrt{\frac{F_{2n+1}}{\phi{F_{2n}}}}$ $n \ge 1$
$F_n$; Fibonacci numbers
$\frac{1+\sqrt5}{2}=\phi$
Prove
$$1+\frac{1}{\sqrt{\phi^{4n+1}{F_{2n}F_{2n+1}}}+\phi^{2n+1}F_{2n}}=\sqrt{\frac{F_{2n+1}}{\phi{F_{2n}}}}$$
I can't go any further can anyone help please?

Expand the equation
Let 
$A=\phi^{4n+1}F_{2n}F_{2n+1}$
$B=\phi^{2n+1}F_{2n}$
$C=\frac{F_{2n+1}}{\phi{F_{2n}}}$
$$\left(1+\frac{1}{\sqrt{A}+B}\right)^2=C$$
$$1+\frac{2}{\sqrt{A}+B}+\frac{1}{(\sqrt{A}+B )^2}=C$$
This method doesn't seem to be working.
Can A, B and C be simplify more further?
I try and rationalise the denominator
$$\frac{2(\sqrt{A}-B)(A-B)+(\sqrt{A}-B)^2}{(A-B^2)^2}=C$$
Still nothing is been simplified.
 [/INST] notice that $A=B^2C$, and since $B>0$, $\sqrt A = B\sqrt C$.
Since $\sqrt A + B > 0$, $1 + 1/(\sqrt A + B) = \sqrt C \iff \sqrt A + B + 1 = \sqrt C(\sqrt A + B)$.
But $\sqrt A+B+1 = B\sqrt C + B + 1$ and $\sqrt C(\sqrt A+B) = BC + B\sqrt C$, so $ \sqrt A + B + 1 = \sqrt C(\sqrt A + B) \iff B+1 = BC \iff \phi^{2n+1}F_{2n}+1 = \phi^{2n}F_{2n+1}$
Now if this is true you should be able to prove it by induction or by using a formula for $F_n$.
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