How do I prove/disprove that $F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1}$, where $F_n$ is the $n$th Fibonacci number?

I was looking at the Fibonacci sequence:

$$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,\ldots, F_n : F_n + F_{n - 1} = F_{n + 1} \land F_0 = 0$$

Here I noticed something; $\qquad F_n^2 + F_{n\pm1}^2 = F_{2n\pm1} : n\in \mathbb{N}$

For example; $$\ \ F_3^2 + F_{3 + 1}^2 = F_3^2 + F_4^2 = 2^2 + 3^2 = F_{2\times3 + 1} = F_{6 + 1} = F_7 = 13$$ $$F_3^2 + F_{3 - 1} = F_3^2 + F_2^2 = 2^2 + 1^2 = F_{2\times3 - 1} = F_{6 - 1} = F_5 = 5$$ However, no matter how true this can seem, I cannot seem to prove/disprove it.

My Attempt: $$\because (a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + b^2 + 2ab : (a,b)\in \mathbb{Z}$$ $$\implies F_n^2 + F_{n\pm 1}^2 = (F_n + F_{n\pm 1})^2 - 2F_nF_{n\pm 1}$$ $$\implies F_n^2 + F_{n - 1}^2 = F_{n + 1}^2 - 2F_nF_{n - 1} : F_n + F_{n - 1} = F_{n + 1}$$ $$\implies F_n^2 + F_{n + 1}^2 = F_{n + 2}^2 - 2F_nF_{n + 1} : F_n + F_{n + 1} = F_{n + 2} \iff F_n + F_{n - 1} = F_{n + 1}$$ $$\implies F_{2n\pm 1} + 2F_nF_{n\pm 1} = F_{n + 2\lor1}^2 : F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1} \tag{+2 if +, +1 if -}$$ $$\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1} - F_{n\pm 1}^2}\sqrt{F_{2n\pm 1} - F_n^2}= F_{n + 2\lor1}$$ $$\implies F_{2n\pm 1} + 2\sqrt{(F_{2n\pm 1} - F_{n\pm 1}^2)(F_{2n\pm 1} - F_n^2)} = F_{n + 2\lor1}^2$$ $$\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1}(F_{2n\pm 1} - F_n^2 - F_{n\pm 1}^2) - (F_nF_{n\pm 1})^2} = F_{n + 2\lor1}^2$$.

$$\because F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1}$$ $$\implies F_{2n\pm 1} + 2\sqrt{-(F_nF_{n\pm 1})^2} = F_{n + 2\lor1} \neq F_{2n\pm 1} + 2\sqrt{\big(-(F_nF_{n\pm 1})\big)^2}$$ $$\therefore F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} : i = \sqrt{-1}$$

I have looked over every step and cannot see where I am wrong. The only part that must be wrong is that:

$$F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} \iff F_n^2 + F_{n\pm1}^2 = F_{2n\pm 1}$$

I can seperate $F_{2n\pm 1}$ as $F_n^2 + F_{n\pm 1}^2$ (if true) so I turn the equation into:

$$F_n^2 + F_{n\pm 1}^2 + 2F_nF_{n\pm 1}i = F_{n + 2\lor1}$$ $$\therefore (F_n + F_{n\pm 1})\bigg(\frac{F_n}{F_{n\pm 1}} + \frac{F_{n\pm 1}}{F_n} + 2i\bigg) = F_{n + 2\lor1}$$ Now I know that the greater the value of $n$, one of the fractions in the above equation (depending on which sign is used in $F_{n\pm 1}$) will get closer to the golden ratio $φ = \frac{1 + \sqrt{5}}{2}$ but $i = \sqrt{-1}$ is not a "real" number, although $F_n$ is.

I then searched for other formulae for finding $F_n$ and came across the one below (being Binet's Formula):

$$F_n = \bigg\{\frac{φ^n - ψ^n}{φ - ψ} : ψ = φ - \frac{\sqrt{20}}{2} = \frac{1 - \sqrt{5}}{2} \ \land \ φ - ψ = \sqrt{5}\bigg\}$$

But I do not know what how to turn that into $F_{n\pm 1}$. Could you please show me where I went wrong (because I must have gone wrong somewhere), and also if you wish, could you please show me how to turn the above fraction into $F_{n\pm 1}$ (if possible).

I only used the $(a + b)^2$ expansion rule because my skill level in mathematics is not particularly high compared to most of the users on this site (especially when logging into the MSE and finding fancy symbols like $\nabla$ and $\partial$ and $\sum$ which all look Greek to me, however the only "fancy" thing I know is that $i = i^5 = \cdots = \sqrt{-1}$) so if you used any other techniques in trying to prove this formula, could you please show me step by step?

Much appreciated. Thank you in advance.

Edit: I believe this question is a duplicate of this other question $\longrightarrow$ Prove $f_{n+1}^2+f_n^2=f_{2n+1}$

and I can go here to serve as a little hint $\longrightarrow$ Fibonacci Square Indentity

Solution 1:

A direct way to prove this is to define $A=\begin{pmatrix}0&1\\1&1\end{pmatrix}$, and prove inductively:

$$\begin{pmatrix}F_{n-1}&F_{n}\\F_{n}&F_{n+1}\end{pmatrix}=A^n$$

Then just equate both sides of $A^{2n}=A^{n}A^{n}$.

This also gives the rule $F_{2n}=F_n\left(F_{n-1}+F_{n+1}\right).$

You can probably prove these two rules inductively together. That is, if:

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

then

$$F_{2n+2}=F_{n+1}(F_{n}+F_{n+2})\\ F_{2n+3}=F_{n+1}^2+F_{n+2}^2$$

For example: $$\begin{align}F_{2n+2}&=F_{2n+1}+F_{2n}\\ &=F_n(F_{n-1}+F_{n+1})+F_{n}^2+F_{n+1}^2\\ &=F_n(2F_{n+1}-F_n)+F_{n}^2+F_{n+1}^2\\ &=2F_nF_{n+1}+F_{n+1}^2\\ &=F_{n+1}(F_n+F_n+F_{n+1})\\ &=F_{n+1}(F_n+F_{n+2}) \end{align}$$

You can do the same for $F_{2n+3}$.