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Let $F(n)$ be $n$-th Fibonacci-number. Then the following holds: $$F(n)^2+F(n+1)^2=F(2n+1).$$ I have proved, but I want a brief proof. Do you know anything about it?

Bram28
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Ray MIZUSAWA
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    I hardly believe there's a shorter proof than induction on this, basically because I don't think this can follow by a combinatorial argument – Michele Caselli May 20 '18 at 12:57

2 Answers2

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I can do a pretty neat proof ... though it does require a bit of set-up.

In particular, it's easily proven by relating the Fibonacci numbers to the following problem:

Suppose that you want to go up some flight of stairs and at every step you can take either one or two stairs: in how many ways can you get up the stairs?

Well, if we say that there are $n$ stairs, then it turns out there are $F_{n+1}$ ways to do it.

A very easy inductive proof shows why:

Base cases: If there is $1$ stairs, you can do it in only $1$ way, and indeed $F_{1+1}=F_2=1$. If there are $2$ stairs, then there are two ways: either take two steps of $1$, or take one step of $2$. And indeed, $F_{2+1}=F_3=2$

Inductive step: Say you have $n >2$ stairs. For your first step you can either go one up or two stairs up. By inductive hypothesis, there are $F_{n}$ ways to finish climbing the $n-1$ stairs after having taken a step of $1$ stairs, and there are $F_{n-1}$ ways to finish climbing the $n-2$ stairs after having taken a step of $2$ stairs. So, there are $F_{n}+F_{n-1}=F_{n+1}$ ways to climb $n$ stairs.

OK, so now that we have made a connection between the Fibonacci numbers and the number of ways to climb stairs in this way, we can prove your desired result very quickly:

Let's climb $2n$ stairs. We now know we can do this in $F_{2n+1}$ ways. But note that there are two different possibilities for us climbing those $2n$ stairs: The first way is that we climb the stairs and that at some point we will have climbed exactly $n$ stairs. If this happens, then there are $F_{n+1}$ ways to climb the first $n$ stairs, and also $F_{n+1}$ ways to climb the other $n$ stairs. The second way is that at some point we will have climbed exactly $n-1$ stairs, which can be done in $F_{n}$ ways, after which we take a step of $2$ stairs, and then finish climbing the remaining $n-1$ stairs, which can be done in $F_{n}$ ways. So, it must be true that:

$$F_{2n+1} = F_{n+1} \cdot F_{n+1} + F_{n} \cdot F_{n}$$

Bram28
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  • Great proof! Minor correction: "We now know we can do this in $F_{2n+1}$ ways" – Jonathan May 20 '18 at 13:27
  • Thank you so much. I think this proof is very wonderful and interesting!! – Ray MIZUSAWA May 20 '18 at 13:35
  • @RayMIZUSAWA I'm glad you like it! Yeah, it makes you wonder what other results involving Fibonacci numbers can be quickly proven with this conceptual connection. :) – Bram28 May 20 '18 at 13:39
  • @Jonathan Glad you like the proof. Thanks for pointing out the typo! – Bram28 May 20 '18 at 14:56
  • HCP16 should read this :) – ama May 20 '18 at 17:50
  • @ama Well, I still used induction to make the connection ... But yes, the induction for that was trivial, while trying to apply induction directly to this problem (as the OP presumably did) would be more laborious. – Bram28 May 20 '18 at 18:01
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My favorite Fibonacci technique is the matrix formulation, which is well worth knowing and easily proved: $$ A^n= \begin{pmatrix}1&1\\1&0\end{pmatrix}^n= \begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix} $$ Therefore, $$ \begin{pmatrix}F_{2n+1}&F_{2n}\\F_{2n}&F_{2n-1}\end{pmatrix} =A^{2n} =(A^n)^2 =\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}^2 =\begin{pmatrix}F_n^2+F_{n+1}^2&*\\*&*\end{pmatrix} $$ The omitted terms give other identities, such as $F_{2n} = (F_{n-1}+F_{n+1})F_n$.

lhf
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