On the equation \(F(n^2+1)=bF(n^2-1)+c\)

Abstract. Under a suitable conjecture concerning prime numbers, we provide a complete solution to the equation $$ F(n^2+1)=bF(n^2-1)+c \quad\text{for every}~n\in {\mathbb N},\ \ n>1, $$ where \(F:{\mathbb N}\to {\mathbb C}\) is a completely multiplicative function and \(b, c\in {\mathbb C},\ c\neq 0\).