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I am trying to understand the proof that "For all $n\geq 2, F_n^2-F_{n+1}F_{n-1}=(-1)^{n-1}$.Where $F_n$ stands for the Fibonacci number at $n$. I got this proof from a book and here is the proof.

Fn+12 = Fn+1 (Fn+Fn-1) by the definition of Fn+1
= Fn+1Fn+ Fn-1Fn+1 + (Fn2 - Fn2)
=Fn2 + Fn+1Fn - (Fn2-Fn-1Fn+1)
=Fn(Fn+Fn+1)-(Fn2-Fn-1Fn+1)
=FnFn+2 - (Fn2 - Fn-1Fn+1) by definition of Fn+2
FnFn+2 -(-1)n-1 by the inductive assumption

Hence Fn+12 - FnFn+2 = (-1)n and the result follows.

I understand most of the proof I just don't understand the last part
FnFn+2 -(-1)n-1 by the inductive assumption
Hence Fn+12 - FnFn+2 = (-1)n and the result follows.
How does this part prove this thoerom? Arent'nt they just assuming that (Fn2 - Fn-1Fn+1) = (-1)n and not proving it?

User1
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  • Wel, first line is wrong. It should be $F_{n+1}^2=F_{n+1}(F_n+F_{n-1})$. – Thomas Andrews Jan 09 '15 at 02:54
  • We're operating under the inductive hypothesis that $F_n^2 - F_{n+1} F_{n-1} = (-1)^{n-2}$ and trying to show that this implies that $F_{n+1}^2 - F_{n+2} F_n = (-1)^{n-1}$. Therefore, they assume the former and prove the latter. As Thomas said, however, the first line of the proof should start with $F_{n+1}^2 = F_{n+1}(F_n + F_{n-1})$. – Clark Zinzow Jan 09 '15 at 02:57
  • OK I just edited the first line – User1 Jan 09 '15 at 02:58
  • @ClarkZinzow So is this not the full proof? – User1 Jan 09 '15 at 02:59
  • No, it appears that proving that the base case for $n=2$ is missing. – Clark Zinzow Jan 09 '15 at 03:01
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    Please use mathjax to format the question. etc is hard to read. http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Aryabhata Jan 09 '15 at 03:01
  • The last line is rather puzzling. From the previous line, we have $F_{n+1}^2-F_{n+2}F_n=-(F_n^2-F_{n+1}F_{n-1})$. By the induction assumption $F_n^2-F_{n+1}F_{n-1}=(-1)^{n-1}$. It follows that $F_{n+1}^2-F_{n+2}F_n=(-1)^n$. – André Nicolas Jan 09 '15 at 03:05
  • I think there is a typo in the "and the result follows" line. The left subscript should be $n+1$ – Ross Millikan Jan 09 '15 at 03:07
  • I tried to fix that, unfortunately it is not an edit over 6 characters long. :( – Clark Zinzow Jan 09 '15 at 03:09
  • Anyways, this is a version of Cassini's Identity, of which two nice proofs (one of which proves it via inductively proving a matrix identity and taking the determinant of both sides! Neat!) exist on ProofWiki: https://www.proofwiki.org/wiki/Cassini's_Identity – Clark Zinzow Jan 09 '15 at 03:12
  • @ClarkZinzow: Got it. – Ross Millikan Jan 09 '15 at 03:24

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I presume the base cases, $n=1$ and $n=2$ were shown by calculation. The idea of induction is to show that if it is true for $n$, it is true for $n+1$. Having shown the base cases, you assume it is true for $n$, and prove it is true for $n+1$. The inductive assumption is $F_n^2-F_{n+1}F_{n-1}=(-1)^{n-2}$ and we want to prove $F_{n+1}^2-F_{n+2}F_{n}=(-1)^{n-1}$ so we are allowed to use the former.

One of my favorites on this subject is Arturo Magidin's answer here

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Ross Millikan
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  • The way I read the proof it was that they just replaced (F_n^2 - F_{n-1}F_{n+1}) with (-1)_{n-1} in lines 5 and 6 of the proof because that was the original assumption. Did I misinterpret what they meant? – User1 Jan 09 '15 at 03:17
  • No, you are exactly correct. That is the assumption that the equation is true for $n$. We are trying to prove it for $n+1$. The argument could be better written to state that assumption at the top. – Ross Millikan Jan 09 '15 at 03:23
  • Wouldn't that cause a sort of circular logic? We assume that (F_n^2 - F_{n-1}F_{n+1}) = (-1){n-1} and then we try to use that fact to prove that (F_n^2 - F{n-1}F_{n+1}) = (-1)_{n-1}. – User1 Jan 09 '15 at 03:28
  • No, that is exactly the point of induction. We assume $(F_n^2 - F_{n-1}F_{n+1}) = (-1)^{n-2}$ and use it to prove $(F_{n+1}^2 - F_{n}F_{n+2}) = (-1)^{n-1}$ where all the indices are increased by one (and your exponents on$-1$ are off by $1$. – Ross Millikan Jan 09 '15 at 03:33
  • The essence of the inductive hypothesis is that we are assuming that the statement $F_n^2 - F_{n-1} F_{n+1} = (-1)^{n-1}$, call it $P(n)$, is true, and using this fact to show that $P(n+1)$, or $F_{n+1}^2 - F_{(n+1)-1} F_{(n+1)+1} = (-1)^{(n+1)-2}$, is true. We wish to show that $P(n) \implies P(n+1)$, so we are allowed to assume that $P(n)$ is true while we judge the validity of $P(n+1)$. – Clark Zinzow Jan 09 '15 at 03:33
  • Does that make sense? I can send you a good number of links about induction, although I am guessing that Ross has a lot more. I'm rather new here. – Clark Zinzow Jan 09 '15 at 03:34
  • Yes but I still don't understand how to go from the sixth line of the proof to the seventh line of the proof. – User1 Jan 09 '15 at 03:37
  • You lost an equal sign on the line that ends "by the inductive assumption". Once you put that in there is a chain of equalities that justifies your last except that the exponent on $-1$ should be $n-2$. That makes the exponent on the last line $n-1$ as required for the proof. – Ross Millikan Jan 09 '15 at 03:47
  • @Ross Millikan OK my confusion more specifically is that they prove through all this algebra that F_{n+1}^2 = F_nF_{n+2} - (-1)^{n-1}. However the book says in the last line of the proof that F_{n+1}^2 - F_nF_{n+2} = (-1)^n. How did they get this line since when you manipulate it algebraically you get F_nF_{n+2}- F_{n+1}^2 = (-1)^{n-1}? So basically how did they make (-1)^{n-1} into (-1)^n and change some of the signs around? – User1 Jan 09 '15 at 20:06
  • @RossMillikan also it should be n-1 and be transformed to n. I made a typo at the beginning and its actually For all n≥2,F^2_n−F_{n+1}F_{n−1}=(−1)^{n−1}. Sorry about that. – User1 Jan 09 '15 at 20:10
  • Between the line ending with "by the inductive assumption" and the next line, they changed $-(-1)^{n-1}$ to $+(-1)^n$. The change of sign reflects the change of one in the exponent. – Ross Millikan Jan 09 '15 at 21:22
  • @Ross Millikan ok thanks I understand it now! – User1 Jan 10 '15 at 17:42