Prove that if $f$ is the Fibonacci sequence then $f_{n+1}^2+f_n^2=f_{2n+1}$ holds for all n. Instead of trying doing this by induction, I need to do it by trying to just replacing the explicit formula for $f_n$. I am stuck at this step:
$f_{n+1}^2+f_n^2=\frac{1}{5}[\phi^{2n}(1+\phi^2)+(1-\phi)^{2n}(1+(1-\phi)^2)]$
Believe me that this is equivalent to $\frac{1}{\sqrt{5}}[\phi^{2n+1}-(1-\phi)^{2n+1}]$
Maybe it's just a simple algebra problem, but I don't know how to get that expression. Thanks
One can show by, say, induction, that: $$A^n=\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n=\begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix}$$ Thus, because $A^{2n}=A^nA^n$, one has that
$$\begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix}\begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix}=\begin{bmatrix} F_{2n+1} & F_{2n} \\ F_{2n} & F_{2n-1} \end{bmatrix}$$
Compare the upper left entry of the right hand matrix, and observe how one can compute it by expanding the left hand product.
The object is to derive the expression $f_{n+1}^2+f_n^2=f_{2n+1}$ from the Binet formula, which I'll express here as
$$f_n=\frac{\varphi^n-\psi^n}{\varphi-\psi}$$
where $\varphi,\psi=(1\pm \sqrt{5})/2$ and, in addition, $\psi=-1/\varphi$.
Expanding the left-hand-side of the expression and rearranging by like terms we obtain
$$f_{n+1}^2+f_{n}^2=\frac{\varphi^{2n}(1+\varphi^2)+\psi^{2n}(1+\psi^2)-2\varphi^n\psi^n(1+\varphi\psi)}{(\varphi-\psi)^2}$$
The derivation hinges on the following three relations...
$$ 1+\varphi^2=\varphi+2=\varphi(\varphi-\psi)\\ 1+\psi^2=\psi+2=-\psi(\varphi-\psi)\\ \varphi\psi=-1 $$
There follows
$$f_{n+1}^2+f_n^2=\frac{\varphi^{2n+1}-\psi^{2n+1}}{\varphi-\psi}=f_{2n+1}$$
Since we've come this far, I took a look at the Lucas numbers, I found that
$$L_{n+1}^2+L_n^2=(\varphi-\psi)^2f_{2n+1}$$
A little bit late, but I was working on this problem for a homework and I think that this proof was not mencionated before.
An alternative proof can by achieved by doble counting. First, we have to show that $F_n$ is the number of ways that you can write $n$ as the sum of 1's or 2's (e.g. $F_3 = |\{ 1+2,1+1+1,2+1\}|$). For this is enough to prove that this is true for $n=1$ and $n=2$, and for $n\geq 3$ notice that if the first term is 1, then the next terms should sum $n-1$ the number of ways of doing this is $F_{n-1}$ and if the first term is 2, then the next terms should sum $n-2$ the number of ways of doing this is $F_{n-2}$. With this facts, we can conclude that the sequence we defined is indeed the fibonacci sequece (if your not conviced you can prove this inductively).
With this in mind, let's consider the number of ways that we can write $2k$ as the sum of 1's or 2's. As we show previously this is $F_{2k}$. Now let's count this in a different way. If by summing term by term, we pass through $k$ then we split the sum into two sums of $k$ by the product principle this is $F_k^2$. If it's imposible to pass through $k$ then we must have the situation, numbers that sum $k-1$, then a 2, then numbers that sum $k-1$. Again, by the product principle this is $F_{k-1}^2$. Therefore, by the sum principle, we have that $$F_{2k} = F_{k-1}^2+F_k^2$$
$\textit{Remark}$: the doble counting argument may have some issues in the case $n=1$, so it must be proven. But this case is trivial because it's a direct result of the definition of Fibonacci numbers.
