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Given $F_m$ be the $m^\text{th}$ number in the Fibonacci sequence. Prove that for all natural $n$, $$\large |F_n^2 + F_nF_{n + 1} - F_{n + 1}^2| = 1$$

(When I'm bored, I do random stuff.)

There has been a solution below if you want to check out. And I would be appreciated if there are other solutions.

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    An interesting related observation is that for positive integers $m,n$ the equation $|m^2+mn-n^2|=1$ implies that $m$ and $n$ are consecutive Fibonacci numbers. This played a key role in an IMO problem in '81. Essentially Vieta jumping. – Jyrki Lahtonen Dec 30 '19 at 06:18
  • $|F_n(F_n+F_{n+1})-F_{n+1}^2|=|F_nF_{n+2}-F_{n+1}^2|=1$ from Cassini's Identity. – mathlove Dec 30 '19 at 06:27
  • For a slightly more advanced connection you should observe that the field norm $N$ of $K=\Bbb{Q}(\sqrt5)$ reads $$N(m+n\frac{1+\sqrt5}2)=(m+n\frac{1+\sqrt5}2)(m+n\frac{1-\sqrt5}2)=m^2+mn-n^2.$$ Therefore the equation I described amounts to find the units of the ring of algebraic integers in $K$. This can also be viewed as solving a Pell equation. Anyway $-1$ and $(1+\sqrt5)/2$ generate that group of units and the claim follows. – Jyrki Lahtonen Dec 30 '19 at 07:53

2 Answers2

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First and foremost, $$F_1^2 + F_1F_2 - F_2^2 = 1^2 - 1 \cdot 1 + 1^2 = 1$$

Assuming that the above statement is correct with $n = p \in \mathbb Z^+$. We have that $$F_{p + 1}^2 + F_{p + 1}F_{p + 2} - F_{p + 2}^2 = F_{p + 1}^2 + F_{p + 1}(F_p+ F_{p + 1}) - (F_p + F_{p + 1})^2$$

$$ = -F_p^2 - F_pF_{p + 1} + F_{p + 1}^2$$

$$ \implies |F_{p + 1}^2 + F_{p + 1}F_{p + 2} - F_{p + 2}^2| = |F_{p + 1}^2 + F_{p + 1}F_{p + 2} - F_{p + 2}^2| = 1$$

Using mathematical induction, for all natural $n$, $|F_n^2 + F_nF_{n + 1} - F_{n + 1}^2| = 1$

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$\begin{array}\\ d_n &=F_n^2 + F_nF_{n + 1} - F_{n + 1}^2\\ &=F_n^2 + F_n(F_n+F_{n - 1}) - (F_n+F_{n - 1})^2\\ &=F_n^2 + F_n^2+F_nF_{n - 1} - (F_n^2+F_{n - 1}^2+2F_nF_{n - 1})\\ &=F_n^2 -F_nF_{n - 1} - F_{n - 1}^2\\ &=-(F_{n - 1}^2+F_nF_{n - 1}-F_n^2)\\ &=-d_{n-1}\\ \end{array} $

$d_1 =F_1^2 + F_1F_{2} - F_2^2 =1+2-4 =-1 $.

Therefore $d_n = (-1)^n$.

marty cohen
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