A real-valued function $f$ has the property that for all real numbers $a$ and $b,$ \[f(a + b)  + f(a - b) = 2f(a) f(b).\] Which one of the following cannot be the value of $f(1)?$
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$
Substituting $a = b$ we get 
\[f(2a) + f(0) = 2f(a)^2\]
Substituting $a= 0$ we find \[2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.\]
This gives \[f(2a) = 2f(a)^2 - f(0) \geq 0-1\]
Plugging in $a = \frac{1}{2}$ implies $f(1) \geq -1$, so answer choice $\boxed{\textbf{(E) -2}}$ is impossible. 
~AtharvNaphade