If $f(x)$ is a real valued function of the real variable $x$, and $f(x)$ is not identically zero, 
and for all $a$ and $b$ $f(a+b)+f(a-b)=2f(a)+2f(b)$, then for all $x$ and $y$
$\textbf{(A) }f(0)=1\qquad \textbf{(B) }f(-x)=-f(x)\qquad \textbf{(C) }f(-x)=f(x)\qquad \\ \textbf{(D) }f(x+y)=f(x)+f(y) \qquad \\ \textbf{(E) }\text{there is a positive real number }T\text{ such that }f(x+T)=f(x)$
We can start by finding the value of $f(0)$. 
Let $a = b = 0$
\[f(0) + f(0) = 2f(0) + 2f(0) \Rrightarrow 2f(0) = 4f(0) \Rrightarrow f(0) = 0\]
Thus, $A$ is not true.
To check $B$, we let $a = 0$. We have
\[f(0+b) + f(0-b) = 2f(0) + 2f(b)\]
\[f(b) + f(-b) = 0 + 2f(b)\]
\[f(-b) = f(b)\]
Thus, $B$ is not true, but $C$ is.
Thus, the answer is $\boxed{C}$
~~JustinLee2017