While following the proof that no Fibonacci number is a perfect square larger than 144 (https://math.la.asu.edu/~checkman/SquareFibonacci.html) I stumbled in proving two of the elementary facts about Lucas and Fibonacci numbers.

(Lucas numbers have the same rule as Fibonacci numbers except that instead of $\{0, 1\}$ the Lucas sequences starst with $\{2,1\}$. The formulas in closed form are $ F_n = \frac{1}{\sqrt{5}}\left[ \rho^n - (-\rho)^{-n}\right] $ and $ F_n = \rho^n + (-\rho)^{-n} $.)

The two properties I got stuck on is are that for $m \in \Bbb{N}$ and $r \in \{1,2\}$, $$(L_{3m},F_{3m}) = 2$$ $$(L_{3m+r},F_{3m+r}) = 1$$

It is easy to show that both $F_n$ and $L_n$ are even only if $n$ is a multiple of 3, but the properties imply there are no other common factors.

What I have tried is to write $ L_s = \frac{p}{q} F_s $, attempting to show that $q$ must be at least $F_s/2$. The starting point is to use the closed form expressions to get $$ \sqrt{5} = \frac{p}{q} +\frac{2}{F_s}(-\rho)^{-s}$$ and use the fact that $(-\rho)^{-s}$ falls rapidly with increasing $s$ so that at some point $\frac{p}{q}$ is "too close" for a rational approximation to $\sqrt{5}$ if $q < \frac{2}{F_s}$. I would then say that for smaller $s$, the properties hold by inspection, and for larger $s$, the value of $q$ must be large enough that the gcd in question cannot exceed 2. But I can't quite get that line of reasoning to work as a proof.


Note that $L_n=F_{n-1}+F_{n+1}$ for all $n$. So $$(F_n,L_n)=(F_n,F_{n-1}+F_{n+1})=(F_n, F_n+2F_{n-1})=(F_n,2F_{n-1}) \leq 2(F_n,F_{n-1})=2 \, .$$

  • $\begingroup$ I like it. Thanks. BTW, are you the Micah who is at IMSA? $\endgroup$ – Mark Fischler Jan 26 '15 at 17:52
  • $\begingroup$ You're quite welcome! I've never been anywhere near IMSA... $\endgroup$ – Micah Jan 26 '15 at 20:12

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