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Prove that for every $ n \in \mathbb{N} \cup \{0\}$, $F_{n+5} + 2F_n$ is divisible by $5$, where $(F_n)_n$ is the Fibonacci sequence.

I think induction won't work here, so I wanted to use a theorem from the lectures, which states: Let $(A_n)$ and $(B_n)$ be arbitrary solutions to the recurrence equation $x_{n+1} = x_n + x_{n-1}$ on $\mathbb{N} \cup \{0\}$ (this means that for every $n \in \mathbb{N}$, $A_{n+1} = A_n + A_{n-1}$ and $B_{n+1} = B_n + B_{n-1}$. If $A_0 = B_0$ and $A_1 = B_1$ (the initial conditions are the same), then it holds that for every $n \in \mathbb{N} \cup \{0\}$, $A_n = B_n$. In other words, they are the same sequence - identical initial conditions uniquely determine the entire sequence).

So I need to somehow cleverly define $A_n$ and $B_n$ and verify the same initial conditions, but due to divisibility, I have no idea how. Thanks for the help.

Kolakoski54
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    If $(F_n)_n$ is the Fibonacci sequence, mention it so that everyone can help you! – Kolakoski54 Jun 23 '24 at 16:04
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    Hello :) Induction works. If $F_{n+5}+2F_n$ is divisible by 5 for all $n\leq m$, then... – Jochen Jun 23 '24 at 16:08
  • Hi, welcome to Math SE. "I think induction won't work here" Hint: prove the following by weak induction: both $F_{n+5}+2F_n$ and $F_{n+6}+2F_{n+1}$ are multiples of $5$. Or you can use strong induction, as @Jochen suggests. – J.G. Jun 23 '24 at 16:08
  • Or you can prove it without any induction – Kolakoski54 Jun 23 '24 at 16:12
  • Why would you discard induction just like that? – jjagmath Jun 23 '24 at 16:15
  • @jjagmath there is not a "unique" solution to this problem. I only state that one of them doesn't need the induction principle! In fact, I'd be glad to see another solution :) – Kolakoski54 Jun 23 '24 at 16:21
  • I would never say that there is a unique solution to this or any other problem. But it seems to me that you discarded a perfectly good method without any reason. – jjagmath Jun 23 '24 at 16:36

2 Answers2

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Only by using (multiple times) $F_{n+2}=F_{n+1}+F_{n}$ for all $n\in\mathbb N$, you get that

$$F_{n+5}+2F_n=5(F_{n+1}+F_{n})=5F_{n+2}$$

which is clearly divisible by $5$.

Kolakoski54
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    Oh right. I just got hint that it's the simplest suing the theorem above. –  Jun 23 '24 at 16:20
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Wanted to find how $2$ was identified.

Using Binet's formula, $F_m=\dfrac{a^m-b^m}{a-b}$ where $a,b$ are the roots of $t^2-t-1=0$

and this, $t^n=F_nt+F_{n-1}$

$$F_{n+5}=\dfrac{a^{n+5}-b^{2n+5}}{a-b}=\dfrac{a^n(F_5a+F_4)-b^n(F_5b+F_4)}{a-b}=F_5F_{n+1}+F_4F_n\equiv\begin{cases}3F_n\pmod5\\2F_{n+1}\pmod3\end{cases}$$

as $F_4=3,F_5=5$

Similarly we can prove $$F_{n+5}=\dfrac{a^{n+5}-b^{2n+5}}{a-b}=\dfrac{a^{n-1}(F_6a+F_5)-b^{n-1}(F_6b+F_5)}{a-b}=F_6F_n-F_5F_{n-1}\equiv\begin{cases}3F_{n-1}\pmod8\\3F_n\pmod5\end{cases}$$ as $F_6=8$

lab bhattacharjee
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