Peter wrote: Prove that no Fibonacci number can be factored into a product of two smaller Fibonacci numbers, each greater than 1. Let us make a contradicting suggestion. Let's suppose that there are three positive integer numbers $a,b,c>2$ so that $F_{a}=F_{b}F_{c}$. We know that $\gcd(F_a;F_b)=F_{\gcd(a,b)}$ then $b|a$. As $a=bm \Rightarrow F_{bm}=F_{b}F_{c}$ and also we should have that $F_c|F_{bm} \Rightarrow c|bm$ then $2c<bm+1$. On the other hand $F_{bm}=F_{b(m-1)+b}=F_{b(m-1)+1}F_{b}+F_{b(m-1)}F_{b-1}$ $\Rightarrow$ $F_{b}F_{c}=F_{bm}>F_{b(m-1)+1}F_{b}$ then $F_{c}>F_{b(m-1)+1}$ so $c>b(m-1)+1$ but we have proved that $2c<bm+1$ thus $bm>2b(m-1)+1 \Rightarrow 2b>bm+1>bm$ so $m=1$.Then $F_c=1$,contradiction!