Quidditch wrote: Let $\{F_n\}^\infty_{n=1}$ be the Fibonacci sequence and let $f$ be a polynomial of degree $1006$ such that $f(k) = F_k$ for all $k \in \{1008, \dots , 2014\}$. Prove that $$233\mid f(2015)+1.$$ Note: $F_1=F_2=1$ and $F_{n+2}=F_{n+1}+F_n$ for all $n\geq 1$. If I am not missing something Then we can construct $f$ as $f(x)=g(x)\prod_{i=1008}^{2014} (x-i)+\sum_{k=1008}^{2014} {\frac{F_k\prod_{i=1008}^{2014} (x-i)}{(x-k)\prod_{i=1008,i\neq k}^{2014}(k-i)}}$ Where $g$ is any polynomial in $R[x]$ But this gives that $deg(f(x))\geq 1007$