Induction Proof for $F_{2n} = F^2_{n+1} - F^2_{n-1}$ As stated in the tag, I'm trying to prove by induction the claim $F_{2n} = F^2_{n+1} - F^2_{n-1}$, where $F_{n}$ is the $n^{th}$ Fibonacci number. I've spent hours on the inductive step without substantial progress, and am hoping someone can provide a path to the desired result.  
To facilitate the proof, we're given $F_{2n-1} = F^2_{n} + F^2_{n-1}$
(Also, in various past attempts not reflected here I tried to use the following substitutions whenever it seemed prudent: $F_{n} = F_{n-1} + F_{n-2} = F_{n+1} - F_{n-1}$; and $(x^{2} - y^{2}) = (x - y)(x + y)$).
For the inductive step, let $n\geq 1$ and assume $F_{2n} = F^2_{n+1} - F^2_{n-1}$. For $n+1$ we want to show $F_{2(n+1)} = F^2_{n+2} - F^2_{n}$. So
\begin{align}
  F_{2(n+1)} &= F_{2n + 2} \\
             &= F_{2n+1} + F_{2n} \\
             &= (F_{2n} + F_{2n-1}) + F_{2n} \\
             &= 2 F_{2n} + F_{2n - 1} \\
             &= 2 (F^2_{n+1} - F^2_{n-1})
                  + F^2_{n} + F^2_{n-1} 
                     &&\text{by the inductive hypothesis} \tag{1} \\
             &=...F^2_{n+2} - F^2_{n}                     
\end{align}
I have two challenges.  The first is squeezing $-2 F^2_{n}$ out of (1).  The only reasonable way I can see to do that is to expand the $-F^2_{n-1}$ term in  $2 F^2_{n+1} - F^2_{n-1}$ to $-(F_{n} - F_{n-2})^2$. If in instead I combine it with $F^2_{n-1}$ then it seems impossible to generate the $-2 F^2_{n}$ term needed to offset $F^2_{n}$ and produce $-F^2_{n}$. 
The second challenge is to find $(F^2_{n+2}$ in (1), and the only way I can see to do that is to expand the $F^2_{n+1}$ term in $2(F^2_{n+1} - F^2_{n-1})$to $(F_{n+2} - F_{n})^2$. Doing that not only generates $2 F^2_{n}$ which totally screws things up, but it also generates a bunch of intermediate terms I can't get rid of, and it's driving me crazy.  
 A: You're given $$f_{2n-1}=f_n^2+f_{n-1}^2\tag1$$ Replace $n$ with $n+1$ to get $$f_{2n+1}=f_{n+1}^2+f_n^2\tag2$$ Subtract (1) from (2) to get $$f_{2n+1}-f_{2n-1}=f_{n+1}^2-f_{n-1}^2\tag3$$ But the left side of (3) is $f_{2n}$. 
A: Method 1. By induction on $n$ we have $F_n=(a^n-b^n)/\sqrt 5$ where $a=(1+\sqrt 5)/2$ and $b=(1-\sqrt 5)/2=-1/a.$ Plug this into your formula. 
Method 2. Let $M$ be the $2\times2$ matrix with top row $(1,1)$ and bottom row $(1,0).$ By induction on $n,$  the top row of $M^n$ is $(F_{n+1},F_n)$ and the bottom row of $M^n$ is $(F_n,F_{n-1}).$ Consider that $M^{2 n} =(M^n)^2$.
Method  3. Your method. For brevity let $$F_{n-1}=a, \quad F_n=b,\quad F_{n+1}=c,\quad F_{n+2}=d.$$ From your 3rd line and the inductive hypothesis, and given $F_{2 n-1}=F_n^2+F_{n-1}^2$, we have $$F_{2 n+2}=2 F_{2 n}+F_{2 n-1}=2 (c^2-  a^2) +(b^2+a^2)=$$ $$=2 c^2+b^2-a^2 =2 (d-b)^2+b^2-(c-b)^2=$$ $$=2(d-b)^2 +b^2-((d-b)-b)^2=d^2-b^2=F_{n+2}^2-F_n^2.$$
A: As this is a difficult result to prove, I shall prove another result and then derive the said result through it.
Proposition: $F_{n+m} = F_n F_{m-1} + F_{n+1} F_m$
Base cases:  $n = 1$:
$\begin{align}
     F_{m+1} &= F_m + F_{m-1} \\
            &= F_2 F_m + F_1 F_{m-1}
  \end{align}$
Case $n = 2$:
$\begin{align}
   F_{m+2} 
     &= F_{m+1} + F_m \\
     &= F_{m-1} + 2 F_m \\
     &= F_2 F_{m-1} + F_3 F_m
\end{align}$
Hypothesis: $F_{k+m-1} = F_{k-1} F_{m-1} + F_k F_m$ and
$F_{k + m}   = F_k F_{m-1} + F_{k +1} F_m$
Induction step: Adding both the hypothesis statements
$\begin{align}
   F_{k + m + 1} &= F_{k+m} + F_{k+m-1} \\
                &=F_{k+1} F_{m-1} + F_{k+1+1} F_m
\end{align}$
Hence, proved.
$F_{n+m} = F_n F_{m-1} + F_{n+1} F_m$
Putting $n = m$:
$\begin{align}
   F_{2n} &= F_n F_{n-1} + F_{n+1} F_n \\
         &= (F_{n+1} - F_{n-1}) F_{n-1} + F_{n+1} F_n \\
         &= F_{n+1} F_{n-1} + F_{n+1} F_n - F^2_{n-1} \\
         &= F_{n+1}(F_{n-1} + F_n) - F^2_{n-1} \\
         &= F^2_{n+1} - F^2_{n-1}
\end{align}$
