, 2017

Solution:
Let $P(x, y)$ be the given assertion. From $P(0,0)$ we get $f(0)^2 = f(0) \Longrightarrow f(0) = 0, 1$.
From $P(x, x)$ we get $f(x)^2 = f(0)$. Thus, if $f(0) = 0$, we have $f(x) = 0$ for all $x$, which satisfies the given constraints. Thus $f(2017) = 0$ is one possibility.

Now suppose $f(0) = 1$. We then have $P(0, y) \Longrightarrow f(-y) = f(y)$, so that $P(x, -y) \Longrightarrow f(x) f(y) = f(x-y) = f(x) f(-y) = f(x+y)$. Thus $f(x-y) = f(x+y)$, and in particular $f(0) = f\left(\frac{x}{2} - \frac{x}{2}\right) = f\left(\frac{x}{2} + \frac{x}{2}\right) = f(x)$. It follows that $f(x) = 1$ for all $x$, which also satisfies all given constraints.

Thus the two possibilities are $f(2017) = 0, 1$.