$7 \cdot 117$ If $f(x)$ is a real-valued function of a real variable $x$, and $f(x)$ is not identically zero, and for all $a$ and $b$,
$$
f(a+b)+f(a-b)=2 f(a)+2 f(b),
$$

then for all $x$ and $y$
(A) $f(0)=1$.
(B) $f(-x)=-f(x)$.
(C) $f(-x)=f(x)$.
(D) $f(x+y)=f(x)+f(y)$.
(E) there is a positive number $T$ such that $f(x+T)=f(x)$.
(28th American High School Mathematics Examination, 1977)
[Solution] Let $a=b=0$ and substitute into the given equation, we have $2 f(0)=4 f(0)$, which means
$$
f(0)=0 \text {. }
$$

Next, let $a=0, b=-x$, and substitute into the given equation, we have
$$
\begin{aligned}
f(x)+f(-x) & =2 f(0)+2 f(-x), \\
f(x) & =f(-x) .
\end{aligned}
$$

Therefore, the answer is $(C)$.