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Let the sequence $\{F_n\}$ satisfy $F_1=a$, $F_2=b$ where $a,b\in\Bbb N_+$, and that $$F_{a+2}=F_{a+1}+F_a~~\forall a\ge1.$$ If $F_x$ is coprime with $F_y$, show that there doesn't exist a term $F_z=F_xF_y$.

The simplest case is $a=b=1$. In this case use the property $F_{(m,n)}=\left(F_m,F_n\right)$ to get $F_{(x,z)}=\left(F_x,F_z\right)$.

This implies that $F_{(x,z)}=F_x$, thus $x\mid z$. Similarly $y\mid z$. Since $F_{(x,y)}=\left(F_x,F_y\right)=1$, we have $(x,y)=1$ or $2$.

So $z\ge\dfrac{xy}2$. $x=1$, $2$ or $3$ clearly impossible. When $4\le x<y$, there's $$F_{\vphantom{\strut}\frac12xy}\ge F_{2y}=F_y\left(F_{y+1}+F_{y-1}\right)>F_xF_y.$$Contradiction!

However I'm far from the problem above.

Umesh Shankar
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    Let's call the original Fibonacci sequence f. Then, $F_n = af_{n-2} + bf_{n-1}$, up to some offsets that you can check for yourself... Will this be useful? Perhaps by applying euclidean algorithm or something, you should be able to reduce it to $(a, b)$. I think :) – Gareth Ma Dec 29 '22 at 14:07

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