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.

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

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}

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?

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

The frequency of remainder 0 suppresses other cases.

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