Induction Proof for $F_{2n} = F^2_{n+1} - F^2_{n-1}$

Question

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.

It is difficult to prove this result on its own. My advice is that you prove a more generalized result, and then derive this as a corollary — Saikat, Jan 25 '16 at 08:25
See also: How to prove this Fibonacci sequence statement: $F^2_{n+1} - F^2_{n-1} = F_{2n}$ — Martin Sleziak, Nov 07 '20 at 11:53

Gerry Myerson · Answer 1 · 2016-02-20T21:57:31.177

6

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 $f_{2n}$.

edited Feb 20 '16 at 21:57

answered Jan 25 '16 at 12:13

Gerry Myerson

185,413

1

Wow, was not expecting something so short and sweet – ASKASK Jan 25 '16 at 12:16
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All the hard work is in proving (1), but since we're told we can just use (1).... – Gerry Myerson Jan 25 '16 at 12:26
That's genius, although I can't use it. The assignment specifies that we prove the claim by induction. Thanks very much! – Scentless Apprentice Jan 25 '16 at 15:05

score 2 · Accepted Answer · edited Feb 20 '16 at 15:47

2

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 $$F_{2 n+2}=2 F_{2 n}+F_{2 n-1}=2 (c^2- a^2) +(b^2+a^2)=$$ $$=2 c^2+b^2-a^2 =2 (d-b)^2+b^2-(c-b)^2=$$ $$=2(d-b)^2 +b^2-((d-b)-b)^2=d^2-b^2=F_{n+2}^2-F_n^2.$$

edited Feb 20 '16 at 15:47

vonbrand

28,394

answered Jan 25 '16 at 09:28

DanielWainfleet

59,529

Method 3 is the solution I need. Thank you so much! Looks like I was pretty close. If I had expanded $(f_{n-1})^{2}$ to $(f_{n+2} - f_{n} - f_{n})^{2})$ I'd have gotten it (although it's doubtful I would have managed to square that term properly). – Scentless Apprentice Jan 25 '16 at 15:10
It also reminded me of something I clearly can't re-learn enough times: simplify complex expressions with variables. – Scentless Apprentice Jan 25 '16 at 15:20
Even when writing on paper I try to change the symbols to $a,b,c,...$ and simplify as much as I can or I tend to lose track of what I'm doing. Although once I tried to show someone how to rationalize the denominator in $1/(\sqrt 2-1)$, When I wrote $1/(\sqrt 2 -1)=(1+\sqrt 2+1)/((\sqrt 2-1)(\sqrt 2+1))$ he insisted that we must simplify by cancelling the upper and lower $(\sqrt 2+1)$. – DanielWainfleet Jan 25 '16 at 21:19

score 2 · Answer 3 · edited Feb 20 '16 at 15:42

2

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 $n = m$:

$\begin{align} F_{2n} &= F_n F_{n-1} + F_{n+1} F_n \\ &= (F_{n+1} - F_{n-1}) F_{n-1} + F_{n+1} F_n \\ &= F_{n+1} F_{n-1} + F_{n+1} F_n - F^2_{n-1} \\ &= F_{n+1}(F_{n-1} + F_n) - F^2_{n-1} \\ &= F^2_{n+1} - F^2_{n-1} \end{align}$

edited Feb 20 '16 at 15:42

vonbrand

28,394

answered Jan 25 '16 at 11:21

Saikat

2,561

1

This is very nice. Thank you! – Scentless Apprentice Jan 25 '16 at 15:05
@ScentlessApprentice Is this acceptable for your assignment ? – Saikat Jan 26 '16 at 00:56
Absolutely, although it's beyond the scope of what I feel I'm capable of right now. I was very close to the solution outlined in Method 3 presented by user254665, but got lost in the profusion of intermediate terms. – Scentless Apprentice Jan 26 '16 at 21:24
@ScentlessApprentice, I hope my edit clears up some... – vonbrand Feb 20 '16 at 15:43
@vonbrand Thanks for the edits. Do you know why the ending lines were coming in a very odd way inspite of there being dollar signs ? – Saikat Feb 20 '16 at 16:02
No clue. MathJax is fickle... and I'm too accustomed to write LaTeX to make such mistakes. – vonbrand Feb 20 '16 at 16:12
Perhaps you should note (it isn't obvious) that you are doing two interleaved inductions, thus two base cases. – vonbrand Feb 20 '16 at 16:13
@von, you really went through all the posts in this thread and changed every $f$ to an $F$? – Gerry Myerson Feb 20 '16 at 21:58
@GerryMyerson, yes. Standard notation for Fibonacci numbers. – vonbrand Feb 20 '16 at 22:07
@vonbrand I'm using strong induction, and require two base cases since I am using two hypothesis statements. – Saikat Feb 21 '16 at 02:13
Hey @vonbrand, thanks for the edits. Hope someone else finds all this great feedback as useful as I did. – Scentless Apprentice Feb 21 '16 at 17:43
@ScentlessApprentice, if you appreciate it, I'm fully paid. Thanks! – vonbrand Feb 21 '16 at 17:44

Induction Proof for $F_{2n} = F^2_{n+1} - F^2_{n-1}$

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