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Let $\mathrm{Fib}_2$ be the set of sums of two Fibonacci numbers: $$\mathrm{Fib}_2 = \{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 18, 21, 22, 23, 24, 26, \dots \}. $$

The elements in $\mathrm{Fib}_2$ have various prime divisors. I found some curious properties about prime divisors.

  1. If we gather remainders of multiples of $79$ divided by $859$, then they seems to be restricted to 9 numbers. That is, $$\{ 79a~\mathrm{mod}~859 : 79a \in \mathrm{Fib}_2 \} = \{ 0, 65, 385, 490, 491, 621, 624, 764, 846 \}.$$

  2. The frequency of remainder $0$ suppresses other cases. For sums of two Fibonacci numbers $< 10^{100}$, the 9 remainders appear as follows: \begin{array}{lllll} 0:879, & 65:72, & 385:36, & 490:72, & 491:72, \\ 621:84, & 624:72, & 764:72, & 846:78 \end{array}

  3. If an element in $\mathrm{Fib}_2$ is divisible by $859$, then it is divisible by $79$.

Another pair is 139 and 461. Consider sums of two Fibonacci numbers $< 10^{100}$. The remainders of multiples of 139 divided by 461 are only two cases: 0 and 322. They appear 1445 times and 210 times, respectively. Also, the multiple of $461$ in $\mathrm{Fib}_2$ is divisible by $139$.

(229, 95419) and (263, 967) are also such pairs.

Are these observations true? That is, do the following conditions hold for those pairs $(a, b)$? Also, how many such pairs exist?

  1. The remainders of multiples of $a$ in $\mathrm{Fib}_2$ divided by $b$ are very restricted.

  2. The frequency of remainder 0 suppresses other cases.

  3. The multiples of $b$ in $\mathrm{Fib}_2$ are divisible by $a$.

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    $\begingroup$ For what it's worth, the sums of two Fibonaccis are tabulated at oeis.org/A084176 But I think the first thing to do is to write down all the remainders of Fibonacci numbers modulo 79. $\endgroup$ – Gerry Myerson May 3 '17 at 13:02
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    $\begingroup$ So these are quite special properties of Fibonacci sequences modulo 79 and 859. (See Pisano period.) In order to check these properties, it suffices to look at all pairs of the first $LCM(\pi(79),\pi(859))$ Fibonacci numbers. It does not explain why and when this happens. Maybe it is just a coincidence: there are many pairs of numbers, and sometimes we get lucky... $\endgroup$ – Ivan Izmestiev May 3 '17 at 13:52
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    $\begingroup$ The Pisano periods $\pi(79)$ and $\pi(859)$ are both $78$. $\endgroup$ – Robert Israel May 3 '17 at 15:35
  • $\begingroup$ @IvanIzmestiev Thank you. I didn't know about Pisano period. It's interesting. $\endgroup$ – P.-S. Park May 4 '17 at 9:25
  • $\begingroup$ Tables of Fibonacci entry points: Part One (pub. The Fibonacci Assciation, Jan 1965) fq.math.ca/Books/Complete/entrypoints1.pdf, p.44 (of the PDF; p.28 of the typed version) is a table of primes $p$ and their entry points $Z(p)$, ordered by $Z(p)$. From that table it can be seen that $Z(79)=Z(859)=78$, etc. p's entry point $Z(p)$ divides, but might be less than, its Pisano period. E.g. $Z(37)=Z(113)=19$ but their Pisano periods are 76. Entry points are still relevant, though, because it says which $F_n$ $p$ divides: $p\mid F(n)\iff Z(p)\mid n$. $\endgroup$ – Rosie F Nov 16 '17 at 21:15

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