Another Fibonacci identity: $F_{2n-1} = F_{n}^2 + F_{n-1}^2$

Question

Here's a problem that is leading me in circles.

Consider the Fibonacci number $F_n$ defined by $F_0 = 0$, $F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for all $n \geq 2$.

Prove that $F_{2n-1} = F_{n}^2 + F_{n-1}^2$.

Tried an induction proof that lead me nowhere fast. So far we have only been given limited identities, and they are all basically summation formulas for the odd terms, or even terms, or all terms up to n. I also though that maybe if it was a difference, the difference of squares could factor to something, but I'm just not getting it.

Answer 1

The identity may be derived from the interesting fact that

$$\left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^k = \left ( \begin{array} \\ F_{k+1} & F_k\\F_k & F_{k-1} \\ \end{array} \right ).$$

From this, we may observe that

$$\begin{align} \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^m \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^n &= \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^{m+n} \\ &= \left ( \begin{array} \\ F_{m+1} & F_m\\F_m & F_{m-1} \\ \end{array} \right ) \left ( \begin{array} \\ F_{n+1} & F_n\\F_n & F_{n-1} \\ \end{array} \right )\\ &= \left ( \begin{array} \\ F_{m+n+1} & F_{m+n}\\F_{m+n} & F_{m+n-1} \\ \end{array} \right ) \end{align} $$

Therefore we can say that, for example,

$$F_{m+n-1} = F_m F_n + F_{m-1} F_{n-1}$$

Plug in $m=n$ and the desired identity follows.

EDIT

The first identity quoted above follows from putting the Fibonacci recurrence into matrix form:

$$\left ( \begin{array} \\F_{k+2} \\ F_{k+1} \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) \left ( \begin{array} \\F_{k+1} \\ F_k \\ \end{array} \right )$$

which may be immediately verified. We may repeat this matrix multiplication $k$ times to get

$$\left ( \begin{array} \\F_{k+2} \\ F_{k+1} \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right )^k \left ( \begin{array} \\F_{2} \\ F_1 \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right )^k \left ( \begin{array} \\1 \\ 1 \\ \end{array} \right )$$

Noting that $F_{k+2}= F_{k+1}+ F_k$ and $ F_{k+1} = F_k + F_{k-1}$, the stated identity follows.

Answer 2

Using Binet's Fibonacci Number Formula,

$F_r=\frac{a^r-b^r}{a-b}$ where $a,b$ are the roots of $x^2-x-1=0\implies ab=-1,a+b=1$

So, $$F_n^2+F_{n-1}^2=\frac{(a^n-b^n)^2+(a^{n-1}-b^{n-1})^2}{(a-b)^2}$$

$$=\frac{a^{2n}+b^{2n}+a^{2n-2}+b^{2n-2}-2(ab)^{n-1}(ab+1)}{(a-b)^2}$$

$$=\frac{a^{2n-1}\left(a+\frac1a\right)+b^{2n-1}\left(b+\frac1a\right)}{(a-b)^2}\text{ as } ab=-1$$

$$=\frac{a^{2n-1}(a-b)+b^{2n-1}(b-a)}{(a-b)^2} \text{ as } \frac1a=-b,\frac1b=-a$$

$$=\frac{a^{2n-1}-b^{2n-1}}{(a-b)}=F_{2n-1}$$

Answer 3

lab bhattacharjee has given excellent direct prove of the identity OP has seeking. I wonder my following indirect prove would count as a prove of the given identity.

If you use Euler's formula for Pythagorean triple, where you can use $n=F_{k-1}$ and $m=F_k$, then the hypotenuse of the triangle will be $c=m^2+n^2=F_k^2 + F_{k-1}^2$, which is equal to $c=F_k^2 + F_{k-1}^2=F_{2k-1}$ as shown in following table:

$$ \begin{array}{c|ccccc} k & n = F_{k-1} & m = F_k & a = F_k^2-F_{k-1}^2 & b = 2F_kF_{k-1} & c = F_k^2+F_{k-1}^2 \\ \hline 3 & 1 & 2 & 3 & 4 & 5 \\ 4 & 2 & 3 & 5 & 12 & 13 \\ 5 & 3 & 5 & 16 & 30 & 34 \\ 6 & 5 & 8 & 39 & 80 & 89 \\ 7 & 8 & 13 & 105 & 208 & 233 \\ k & F_{k-1} & F_{k} & F_{k-2}F_{k+1} & 2F_kF_{k-1} & F_{2k-1} \\ \hline \end{array} $$

Close inspection of above table reveals that $c = F_k^2+F_{k-1}^2 = F_{2k-1}$. For example, for the first line, $k=3$ so that $F_{2k-1}= F_{2\cdot3-1} = F_5=5$. For the second line, $k=4$ so that $F_{2k-1}= F_{2\cdot4-1} = F_7=13$.

Interestingly, close inspection of above table also reveals that $a = F_k^2-F_{k-1}^2 = F_{k-2}F_{k+1}$, which is also a Fibonacci identity.

Answer 4

Let me show an interesting, direct algebraic proof of your identity involving the Lucas numbers. The Lucas numbers are defined as

$$L_n = F_{n+1} + F_{n-1}$$

For the proof you need the following four results (all proven inductively): \begin{equation} F_1F_2 + F_2F_3 + ... + F_{2n-1}F_{2n} = F_{2n}^2 \quad \quad(1) \end{equation} \begin{equation} F_1F_2 + F_2F_3 + ... + F_{2n-1}F_{2n} + F_{2n}F_{2n+1} = F_{2n+1}^2 - 1 \quad \quad(2) \end{equation} \begin{equation} F_2 + F_4 + F_6+... + F_{2n} = F_{2n+1} -1 \quad \quad (3) \end{equation} \begin{equation} F_{2n} = F_{n}L_{n} \quad \quad(4) \end{equation}

From (1) and (2) we can write \begin{align*} F_n^2+F_{n-1}^2 &= 1 + (F_0F_1 + F_1F_2) + (F_2F_3 + F_3F_4) + ...(F_{n-2}F_{n-1} + F_{n-1}F_n) \\ &= 1 + F_1(F_0+F_2) + ... + F_{n-1}(F_{n-2} + F_{n}) \\ &= 1 + F_1L_1 + F_2L_2 +... + F_{n-1}L_{n-1} \end{align*} Now using (3) and (4) we can write \begin{align*} F_n^2+F_{n-1}^2 &= 1 + F_2 + F_4 ... + F_{2n-2} \\ &= 1 + [F_{2n-1} -1] \\ &= F_{2n-1} \end{align*}