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Given $a,b\in\mathbb Z^+$, and let $F_{a,b}:\mathbb N\to\mathbb N$ be a function such that $F_{a,b}(0)=0$, $F_{a,b}(1)=1$ and $F_{a,b}(n+1)=a\cdot F_{a,b}(n)+b\cdot F_{a,b}(n-1)$. $F_{1,1}$ correspond to the Fibonacci function.

Conjecture for which I would like to see a proof:

If $a,b$ are co-prime and $p>2$ is a prime that doesn't divide $b$, then

$F_{a,b}\left(p-\left(\frac{a^2+4b}{p}\right)\right)\equiv 0\pmod p$,

where $\left(\frac{a^2+4b}{p}\right)$ is the Legendre symbol.

I have tested the conjecture with $10,000$ randomly selected $a,b,p$.

Observation: for Fibonacci numbers $a=b=1$ and $a^2+4b=5$.
Wikipedia

See also The $\gcd$ operator commutes with functions defined by linear recurrence relations

I've never been interested in the Fibonacci numbers and when I recently read about them on Wikipedia I realized that all this interesting theorems for Fibonacci numbers could be generalized to linear recurrence relations with constant coefficients.

This funny relation $p|F_{p-(\frac{5}{p})}$ promted for an explanation. I guessed that the '$5$' came from the solution of the characteristic polynomial of the recurrence relation for the Fibonacci relation with the solutions $\frac{1\pm\sqrt{5}}{2}$. Subtracting the greater root from the smaller and square gives $5$ and this seems to be correct. For the general case $x_{n+2}-ax_{n+1}-bx_n=0$ the characteristic polynomial is $t^2-at-b=0$ with the solutions $\frac{a\pm\sqrt{a^2+4b}}{2}$.

Since I have an excellent system to check hypotheses (BigZ) the conjecture was easy to test computationally.

1 Answers 1

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I'm running low on time so here is half:

Suppose $a^2 + 4b$ is a quadratic residue mod $p$, then indeed we can see by completing the square $x^2 - ax -b$ has two solutions in $\mathbb{F}_p$ lets call them $c,d$.

It's then easy to show by induction that mod $p$ we have that $F_{a,b}(n) \equiv \frac{1}{c-d}c^n - \frac{1}{c-d} d^n$

But then Fermat's little theorem tells us that $c^{p-1} \equiv 1$ and $d^{p-1} \equiv 1$ so $F_{a,b}(p-1) \equiv 0$ as desired.

The other case is more complicated and I don't have time to write it right now but you can take the same approach as above but this time $c,d$ lie in $\mathbb{F}_{p^2}$, the quadratic field extension of $\mathbb{F}_p$.