2

I was wondering if the following is already a known result in mathematics. I have tested it and it seems to work every single time.

If I write the Fibonacci sequence in $\bmod (a)$ form and it repeats after $b$ terms, I will call this the period, and one of the two conditions is true then $a$ is a prime.

If $$b = \frac{a - 1}{n}$$ such that $\frac{a - 1}{n}$ yields a natural number.

Or:

If $b = n(a + 1)$

Then $a$ must be a prime number.

However if $a$ is a prime will not necessarily exhibit these properties.

I would really appreciate your help.

Robert Soupe
  • 14,663
Matthew
  • 21
  • 1
    You may be interested in http://math.arizona.edu/~ura-reports/071/Campbell.Charles/Final.pdf. I've also verified your result for all integers $a \leq 500$. – Patrick Stevens Aug 21 '15 at 11:30
  • 1
    Star from Wikipedia. Also see this older thread (the title is only about modulo 3, but the discussion is more general). Also observe that whether $a+1$ or $a-1$ appears (your 2nd or 1st cases) depends on whether the prime is a quadratic residue modulo 5 or not. This is a consequence of Binet's formula (and quadratic reciprocity). If five is a non-residue, we need to use the field $\Bbb{F}_{p^2}$ to find a counterpart to $\sqrt 5$. – Jyrki Lahtonen Aug 21 '15 at 11:57
  • 1
    See this answer of yours truly for an example, where there is no $\sqrt 5$ modulo $p$ (there $p=503$). Not exactly your question but related. Here's why. By Binet's formula you are interested in the period of powers of the golden ratio modulo $a$. If $a\neq5$ is a prime that period is a factor of $a-1$ when $a$ is a prime such that $5$ is a quadratic residue modulo $a$, and a multiple of $a+1$ otherwise. If $a$ is a non-prime, then it's more complicated. In the former case it will be a factor of $\phi(a)$ instead of $a-1$. But that's not all. – Jyrki Lahtonen Aug 21 '15 at 12:18
  • 1
    The Fibonacci numbers have been studied extensively, and anything you can discover about them is probably already in the book Fibonacci and Lucas Numbers with Applications by Thomas Koshy. However, Koshy might approach your result from a slightly different angle, or it might not even be obvious. – Robert Soupe Aug 21 '15 at 14:52

0 Answers0