WakeUp wrote: Consider the functions $f$ defined on the set of integers such that \[f(x)=f(x^2+x+1)\] for all integer $x$. Find $(a)$ all even functions, $(b)$ all odd functions of this kind. $f(x-1)=f(x^2-x+1)=f(-x)$ (a) If $f(x)$ is even, this implies $f(x-1)=f(x)$ and so $\boxed{f(x)=c}$, constant, which indeed is a solution. (b) If $f(x)$ is odd, this implies $f(x-1)=-f(x)$ and so the function is $f(2n)=a$ and $f(2n+1)=-a$ for some $a$. And since odd, this implies $f(0)=0$ and so $a=0$ and $\boxed{f(x)=0}$ $\forall x$, which indeed is a solution.