# Fiboncacci theorem: Proof by induction

I have the following theorem to prove by induction:

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

It is mentioned in my script that the proof should be possible only by using the recursion of Fibonacci numbers:$$F_{n+1}=F_{n}+F_{n-1}$$

I was only able to proof this theorem by using other theorems, as I was not able to proof it without knowing an relationship between $F_{n }$ and $F_{2n}$. The thing is: I know that the theorem is true, and I have 2 different proofs for it, but I am searching for an elegant proof without using other knowledge apart from the recursion.

Induction hypothesis: $$F_{n} \cdot F_{n+1} - F_{n-2}\cdot F_{n-1}=F_{2n-1}$$ We want to show the following is also true:

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

So we transform the induction hypothesis by adding $F_{n+1}^2-F_{n-1}^2$:

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

So we have the LHS of what we want to show. We would need to show that $F_{n+1}^2-F_{n-1}^2=F_{2n}$ and our proof would be finished. I was not able to proof this, I tried it with another induction.

Another idea was to use two induction hypothesis: $$(1) F_{n-1} \cdot F_{n} - F_{n-3}\cdot F_{n-2}=F_{2n-3}$$ $$(2) F_{n} \cdot F_{n+1} - F_{n-2}\cdot F_{n-1}=F_{2n-1}$$

and then $(2)-(1)= F_{n-2}$ would give us: $$F_{2n-2}= F_{n}^2-F_{n-2}^2$$ This looks more promising, considering our problem in the last attempt. By adding (2) we get: $$F_{2n}= F_{n} \cdot F_{n+2} - F_{n-2}\cdot F_{n}$$ By adding (2) again we would get: $$F_{2n+1}= F_{n} \cdot F_{n+3} - F_{n-2}\cdot F_{n+1}$$ Now one would need to show this is equivalent, to what we want to prove. Again I was not successful trying to do this.

I would be very glad if someone could show me an complete induction proof not requiring other theorems. I know this theorem is only a special case of the following: $$F_{n+m} = F_{n−1}\cdot F_{m} + F_{n}\cdot F_{m+1}$$, which in fact is easy to show with induction. But I only know this theorem from googeling and I don't think it is necessary to do an induction proof of my special case of the theorem only by assuming the general case.

I hope someone would like to help me!

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Do you have $F_0 = F_1 = 1$ or $F_1 = F_2 = 1$? –  Arthur May 4 '13 at 12:13
I have $F_0=0$ and $F_1=1$ whichs means $F_1=F_2=1$ –  GEO May 4 '13 at 12:20

You can apply induction to $F_{2n}$ simultaneously:
\begin{align} F_{2n} &= F_{2n-1}+F_{2n-2} \\ &=F_nF_{n+1}-F_{n-2}F_{n-1}+F_n^2-F_{n-2}^2 \\ &=F_nF_{n+2}-F_nF_{n-2} \\ &= F_n(F_{n+2}-F_{n-2}) \\ &= (F_{n+1}-F_{n-1})(F_{n+1}+F_{n-1}) \\ &=F_{n+1}^2-F_{n-1}^2\;. \end{align}
The following reasoning proves $F_{2n} = F_{n+1}^2-F_{n-1}^2$, which would prove your first induction hypothesis
$$F_{2n} = F_{2n-1} + F_{2n-2} = 2F_{2n-2} + F_{2n-3} = 3F_{2n-3} + 2F_{2n-4}=5F_{2n-4} + 3F_{2n-5} \\\\ \vdots\\\\ = F_{i+1}\cdot F_{2n-i} + F_{i}\cdot F_{2n-i-1}$$ Specifically, for $i=n-1$, we have that $F_{2n} = F_{n}F_{n+1} + F_{n}F_{n-1}$. We now only need to check that $$F_{n}F_{n+1} + F_{n}F_{n-1} = (F_{n+1}-F_{n-1})F_{n+1} + F_{n}F_{n-1} = F_{n+1}^2 - F_{n+1}F_{n-1} + F_nF_{n-1} \\\\ = F_{n+1}^2 - (F_n + F_{n-1})F_{n-1} + F_nF_{n-1} = F_{n+1}^2 - F_{n-1}^2$$ and we are done.