Here,
$F(0) = 0, F(1) = 1,$ and $F(n) = F(n − 1) + F(n − 2)$ for $n ≥ 2.$
Please give hints, not full answers.
Thanks!
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1Look at the divisors of $2013$. But first try with a smaller number. – Daniel Fischer Oct 27 '16 at 20:29
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@DanielFischer Here, we have $gcd(2^{11}-1,F(2013))=89$, but what can we do in general ? – Peter Oct 27 '16 at 20:32
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$F(n)$ is the $n$-th Fibonacci number. – E. Joseph Oct 27 '16 at 20:32
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@Peter What do you know about $F(n)$ and $F(kn)$? – Daniel Fischer Oct 27 '16 at 20:33
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1OK, I see it. $F(n)|F(kn)$ , right ? So, we simply take a proper divisor $d$ of $n$ and calculate the prime factors of $F(d)$ , which are also prime factors of $F(n)$. – Peter Oct 27 '16 at 20:34
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@Peter how are you getting $F(n) | F(kn)$ ? – uuuuuuuuuu Oct 27 '16 at 20:38
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@Saikat I do not know how to prove it, I only noticed that it solves the problem. – Peter Oct 27 '16 at 20:51
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@Saikat One way is induction. Another way is to use Binet's formula and the corresponding formula for the Lucas numbers. – Daniel Fischer Oct 27 '16 at 21:00
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For instance, $$ 89 \mid F_{11} \mid F_{2013}. $$ – Jack D'Aurizio Oct 27 '16 at 21:02