Talk:GCD of Consecutive Integers of General Fibonacci Sequence

Stronger Result
I believe I have proven that instead of $f \divides d$ in the conclusion, we can get the stronger result $f = d$.

The induction proof shown on the page is one evidence.

The following argument also shows that $f = d$ should be correct.

Suppose the weaker theorem is true.

Then we choose an arbitrary $\FF = \sequence {a_n}$ according to the assumptions of the theorem.

Let $\sequence {b_n}$ be defined by:
 * $b_n = \dfrac {a_n} d$

This sequence is essentially the same generalized Fibonacci sequence, with $b_0 = \dfrac r d$, $b_1 = \dfrac s d$, and $t, u$ remains the same.

$\gcd \set {b_0, b_1} = 1$ by Integers Divided by GCD are Coprime.

By the theorem, for any $m$, if $f_b = \gcd \set {b_m, b_{m - 1}}$ and $\gcd \set {f_b, t} = 1$, we would have:
 * $f_b \divides \gcd \set {b_0, b_1} = 1$

so $f_b = 1$.

Now let $f_a = \gcd \set {a_m, a_{m - 1}} = \gcd \set {d b_m, d b_{m - 1}} = d \gcd \set {b_m, b_{m - 1}} = d f_b$ and suppose $\gcd \set {f_a, t} = 1$.

By Divisor of One of Coprime Numbers is Coprime to Other, $\gcd \set {f_a,t} = 1$ implies $\gcd \set {f_b,t} = 1$.

We apply the theorem to give $f_b = 1$.

This shows that $\gcd \set {a_m, a_{m - 1}} = d$, which implies the stronger form of the theorem.

--RandomUndergrad (talk) 11:43, 1 March 2022 (UTC)


 * Hmm ... interesting ... wonder if it's too late to ask what he was thinking of when he crafted this exercise. He's only $83$. --prime mover (talk) 20:53, 1 March 2022 (UTC)