amparvardi wrote: Let $\{fn\}$ be the Fibonacci sequence $\{1, 1, 2, 3, 5, \dots.\}. $ (a) Find all pairs $(a, b)$ of real numbers such that for each $n$, $af_n +bf_{n+1}$ is a member of the sequence. Lemma : If $f_a=f_b+f_c$, then $(a,b,c)\in\{(n+2,n+1,n),(n+2,n,n+1),(4,3,1),(4,1,3)\}$ for some positive integer $n$

$af_n+bf_{n+1}\in\{f_i\}$ $\forall n$. So : $n=1$ $\implies$ $a+b=f_p$ $n=2$ $\implies$ $a+2b=f_q$ $n=3$ $\implies$ $2a+3b=f_r$ $n=4$ $\implies$ $3a+5b=f_s$ So $f_p+f_q=f_r$ and $f_q+f_r=f_s$ And so, using lemma : $(r,p,q)\in\{(n+2,n+1,n),(n+2,n,n+1),(4,3,1),(4,1,3)\}$ and so $(q,r)\in\{(n,n+2),(n+1,n+2),(1,4),(3,4)\}$ for some $n$ $(s,q,r)\in\{(m+2,m+1,m),(m+2,m,m+1),(4,3,1),(4,1,3)\}$ and so $(q,r)\in\{{(m+1,m),(m,m+1),(3,1),(1,3)\}\}}$ for some $m$ So $(p,q,r,s)=(p,p+1,p+2,p+3)$ So $a=2f_p-f_{p+1}$ and $b=f_{p+1}-f_p$ Then : $p=1$ $\implies$ $(a,b)=(1,0)$ which indeed is a solution since $af_n+bf_{n+1}=f_n\in\{f_i\}$ $\forall n$ $p=2$ $\implies$ $(a,b)=(0,1)$ which indeed is a solution since $af_n+bf_{n+1}=f_{n+1}\in\{f_i\}$ $\forall n$ $p>2$ $\implies$ $(a,b)=(f_{p-2},f_{p-1})$ which indeed is a solution since $af_n+bf_{n+1}=f_{p-2}f_n+f_{p-1}f_{n+1}=f_{p+n-1}\in\{f_i\}$ $\forall n$ Hence the answer : $\boxed{(a,b)\in\{(0,1),(1,0)\}\cup\left(\bigcup_{k\in\mathbb N}\{(f_k,f_{k+1})\}\right)}$

For b),the only answer is $u=v=1$.