$b|1-2a$ $b |f_2-2*2a^2=1-4a^2$ $b|f_4-2*8a^4=3-8a^4$ $b| 2*(3-8a^4)-(1-4a^2)(1+4a^2)=6-16a^4-(1-16a^4)=5$ $b=5$ $5|1-2a \to a=3$ We will prove by induction, that $5 | f_n-2n3^n$. For $n=1,2$ it is true. Let it is true for $1,2,...n+1$ $f_{n+2}-2(n+2)3^{n+2}=f_{n}-2n3^n+f_{n+1}-2(n+1)3^{n+1}+2n3^n+2(n+1)3^{n+1}-2(n+2)3^{n+2} = f_{n}-2n3^n+f_{n+1}-2(n+1)3^{n+1}+2 *3^n(n+3(n+1)-9(n+2))= f_{n}-2n3^n+f_{n+1}-2(n+1)3^{n+1}+2 *3^n(-5n-15) \equiv 0 \text{mod} 5$ So $ (a,b)=(3,5)$