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 $f(2017)=0,1$.