# Identity of Fibonacci sequence

Let $F_n$ be a Fibonacci sequence with initial terms $F_0=0, F_1=1$ and $F_{n+1}=F_n+F_{n-1}$ for $n\geqslant 1$.

Prove that $F_n^2+F_{n+1}^2=F_{2n+1}$ for $n\geqslant 0$ (with mathematical induction).

My efforts: For $n=0$ it is true.

Suppose that our statement holds for $0\leqslant k \leqslant n$ i.e. $F_k^2+F_{k+1}^2=F_{2k+1}$

Let's try to prove it for $k=n+1$. $$F_{2n+3}=F_{2n+1}+F_{2n+2}=F_{n+1}^2+(F_{n}^2+F_{2n+2})= ?$$

Here I'm stuck and I have applied different methods but none of them brings a positive result.

Can anyone help to complete this?

• math.stackexchange.com/questions/295173/… – lab bhattacharjee Nov 28 '17 at 11:14
• @labbhattacharjee, thanks but your link has matrix proof which is given below and proof suing binet's formula. As I said above i am interested in induction proof – ZFR Nov 28 '17 at 11:18

This follows from the matrix formulation, which is well worth knowing and easily proved: $$\begin{pmatrix}1&1\\1&0\end{pmatrix}^n= \begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}$$ Just compare $$\begin{pmatrix}1&1\\1&0\end{pmatrix}^{2n}= \begin{pmatrix}F_{2n+1}&*\\*&*\end{pmatrix}$$ with $$\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}^2= \begin{pmatrix}\cdots&*\\*&*\end{pmatrix}$$ Alternatively (and equivalently), you can use induction to prove simultaneously that $$F_{2n} = F_n (F_{n+1} + F_{n-1}), \quad F_{2n+1}=F_n^2+F_{n+1}^2$$

• Adapted from math.stackexchange.com/a/1505284/589 – lhf Nov 28 '17 at 11:14
• Nice proof but I would like to see the induction proof – ZFR Nov 28 '17 at 11:15
• How to use induction in order to prove the second identity? Could you write the induction proof? – ZFR Nov 28 '17 at 11:34
• @RFZ, proceed as in lab bhattacharjee's answer. – lhf Nov 28 '17 at 12:37
• I have added an answer. Please take a look – ZFR Nov 28 '17 at 15:21

We can prove more general identity, i.e. $F_{n+m+1}=F_{m+1}F_{n+1}+F_{m}F_{n}$.

Let's prove above by mathematical induction on $n$.

For $n=0$ we have $F_{m+1}=F_{m+1}$

Suppose it is true for all $0\leqslant n \leqslant k$.

Let's prove it for $n=k+1$ then $$F_{k+m+2}=F_{k+m}+F_{k+m+1}=(F_{m+1}F_{k}+F_{m}F_{k-1})+(F_{m+1}F_{k+1}+F_{m}F_{k})=F_{m+1}(F_k+F_{k+1})+F_m(F_{k-1}+F_k)=F_{m+1}F_{k+2}+F_mF_{k+1}$$

The first two parentheses were derived from induction assumption for $n=k-1$ or $n=k$.

Thus, we have proved our initial statement. If we put $n=m$ we get $$F_{2n+1}=F_{n+1}^2+F_n^2$$

$$F_{2(n+1)+1}=F_{2n+1}+F_{2n+2}$$

$$=F_{n+1}^2+(F_{n}^2+F_{2n+2})$$

$=F_{n+1}^2+F_{n}^2+F_{n+1}(F_{n+2}+F_n)$ (Using Proving a Fibonacci identity: $F_{2n} = F_n (F_{n+1} + F_{n-1})$,)

$$=F_{n+1}^2+F_{n+1}F_{n+2}+F_n(\underbrace{F_n+F_{n+1}})$$

$$=F_{n+1}^2+F_{n+1}F_{n+2}+F_nF_{n+2}$$

$$=F_{n+1}^2+F_{n+2}(\underbrace{F_{n+1}+F_n})$$

$$=?$$