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Let $(a_n)_{n \in \mathbb{N}}$ be the sequence of integers defined recursively by $a_0 = 0$, $a_1 = 1$, and $a_{n+2} = 4a_{n+1}+a_{n}$ for $n \geq 0$. Prove that if $\gcd(a_m,a_{m+1}) = d > 1$, then $\gcd(a_m,a_{m-1}) = d$ where $m \geq 2$.

It can be shown that if $m|n$, then $a_{m}|a_{n}$. I am wondering if this helps prove the statement or we have to use the Euclidean algorithm to derive it.

Majid
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user19405892
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2 Answers2

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Rewrite the recurrence as $$ a_{m+1} = 4a_m +a_{m-1} $$ Now suppose that $d$ divides both terms on the right. Can you explain why it divides the left?

John Hughes
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Assuming $\gcd(a_m,a_{m-1}) = d$, we have $d|a_m$ and $d|a_{m-1}$. Moreover, according to $a_{n+2} = 4a_{n+1}+a_{n}$, we have $a_{m+1} = 4a_m +a_{m-1}$. So, it is concluded that:

$$d|a_m\ \ \ \&\ \ \ d|a_{m-1} \rightarrow \ \ d|(4a_m +a_{m-1})\to \ d|a_{m+1} \to \ d|\gcd(a_m,a_{m+1})$$

Now, assuming $\gcd(a_m,a_{m+1})=d'$, according to $a_{m-1} = -4a_m +a_{m+1}$, similarly it is concluded that $d'|\gcd(a_m,a_{m-1})=d$.

Thus, we can conclude that $d=d'$.

Majid
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  • OP wants to avoid the Euclidean algorithm, but you are essentially using that, i.e. you are using $\ \color{#c00}{a_{m+1}\equiv a_{n-1}} \pmod{a_m}\ $ so $\ (\color{#c00}{a_{m+1}},a_m) = (\color{#c00}{a_{m-1}},a_n),$ by Euclid. Your asnwer just repeats the proof of this gcd equality in this special case. – Bill Dubuque Jul 22 '16 at 14:10