RagvaloD wrote: $f(x)=ax^2+bx+c;a,b,c$ are reals. $M=\{f(2n)|n \text{ is integer}\},N=\{f(2n+1)|n \text{ is integer}\}$ Prove that $M=N$ or $M \cap N = \O$ if $t=\frac {-b}{a}$ prove that $f(n)=f(m) \iff n-t=t-m \iff 2t=n+m$; then you can easily see that if there exists $n,m \in \mathbb {Z}$ such that $f(2n)=f(2m+1)$ then $f(2k)=f(2(n-k+m)+1)$.