Prove a function is constant function with the given conditions. 
If $f(4-x)=f (4+x)$ and $f(2-x)=f(2+x)$, prove that $f$ is constant function. 

I tried to solve by assuming x and y are two variable then if we get $f (x)=f (y)$  then $f$ is constant function, but can't complete the problem. Someone please help me.
 A: $f$ is not necessarily constant. Here is a counter-example:
$$
  f(x) =
\begin{cases}
0,  & \text{if $x\in\Bbb Q$} \\
1, & \text{if $x\notin\Bbb Q$}
\end{cases}$$
And another one, continuous this time:
$$f(x)=\cos\pi x$$
A: The two equations $\,f(4-x)=f(4+x)\,$ and $\,f(2-x)=f(2+x)\,$ require that the function is invariant with respect to two reflections and hence a translation by $4$. The simple calculation is $\, x \mapsto 8-x \,$ in the first reflection and then $ 8-x \mapsto x-4 \,$ after the second reflection yielding a translation by $4.$ Assuming that the function is defined for all real numbers, and that no other conditions apply, then you can let the function be arbitrarily defined on the interval $\,[0,2].\,$ 
The function must be an even function with a period of $4$. That is, $\,f(x)=f(-x)\,$ and $\,f(x)=f(x+4)\,$ for all real $\,x.\,$ This determines the function for all real numbers and that it satisfies $\,f(2 n-x)=f(2 n+x)=f(x)\,$ and $\, f(4 n + x) = f(x) \,$ for all integer $\,n\,$ and all real $\,x.\,$
For a simple example, consider $\, f(4n+x) := |x|\,$ for all $\,|x|\le 2\,$ and integer $\,n.\,$ This is a sawtooth function with fundamental period $\,4.\,$
A: Without tellings what are the domain and the codomain of $f$, nobody can answer that. However, if the domain is, say, $\mathbb Q$, then the statement is false. Consider, for instance,$$\begin{array}{rccc}f\colon&\mathbb Q&\longrightarrow&\mathbb Z\\&q&\mapsto&\begin{cases}1&\text{ if }q\in\mathbb Z\\0&\text{ otherwise.}\end{cases}\end{array}$$
