<s>[INST] Show that $f(2n)= f(n+1)^2 - f(n-1)^2$ Let $f(n)=f(n-1)+f(n-2)$ be the Fibonacci sequence with $f(0)=0,f(1)=1$. Show that $$f(2n)= f(n+1)^2 - f(n-1)^2.$$
I have tried several different approaches to this problem. Both inducting from the rhs and the lhs, and I end up going in circles. Any help would be appreciated! Thanks!
Working from the lhs I've gotten:
$$f(2(n+1)+1)=f(2n+1)+f(2n)$$
$$=f(2n-1) + f(2n) + f(2n)$$
$$=f(2n-1)+2(f(n+1)^2-f(n-1)^2)$$
But I can't figure out where to go from here. Help?
 [/INST] Following up on Joseph's suggestion, we can prove the identity directly using Binet's formula
$$f(n) = \frac{\varphi^n - \psi^n}{\sqrt{5}}$$
where
$$\varphi = \frac{1 + \sqrt{5}}{2} \qquad \psi = \frac{1 - \sqrt{5}}{2} = -\frac{1}{\varphi}$$
so that
$$\varphi^2 - \frac{1}{\varphi^2} = \sqrt{5} = \frac{1}{\psi^2} - \psi^2 \qquad \varphi\psi - \frac{1}{\varphi\psi} = 0.$$
Then
\begin{align}
f(n+1)^2 - f(n-1)^2 & = \left(\frac{\varphi^{n+1} - \psi^{n+1}}{\sqrt{5}}\right)^2 - \left(\frac{\varphi^{n-1} - \psi^{n-1}}{\sqrt{5}}\right)^2\\
& = \frac{1}{5}\left(\varphi^{2n+2} - 2\varphi^{n+1}\psi^{n+1} + \psi^{2n+2} - \varphi^{2n-2} + 2\varphi^{n-1}\psi^{n-1} - \psi^{2n-2}\right)\\
& = \frac{1}{5}\left(\left(\varphi^2 - \frac{1}{\varphi^2}\right)\varphi^{2n} - 2\left(\varphi\psi - \frac{1}{\varphi\psi}\right)\varphi^n\psi^n - \left(\frac{1}{\psi^2} - \psi^2\right) \psi^{2n}\right)\\
& = \frac{1}{5}\left(\sqrt{5}\varphi^{2n} - 2 \cdot 0 \cdot \varphi^n\psi^n - \sqrt{5}\psi^{2n}\right)\\
& = \frac{\varphi^{2n} - \psi^{2n}}{\sqrt{5}}\\
& = f(2n).
\end{align}
</s>