jmc

Setting $x=y=0,$ we get $$f(0{)}^2=f(0),$$so $f(0)=0$ or $f(0)=1.$

Suppose $f(0)=0.$ Setting $y=0,$ we get $$f(x)f(0)=f(x),$$so $f(x)=0$ for all $x.$ Note that this function works, and in particular, $f(2017)=0.$

Now suppose $f(0)=1.$ Setting $x=0,$ we get $$f(0)f(y)=f(-y),$$so $f(-y)=f(y)$ for all $y.$

Replacing $y$ with $-y$, we get $$f(x)f(-y)=f(x+y).$$Then $f(x+y)=f(x)f(-y)=f(x)f(y)=f(x-y)$ for all $x$ and $y.$ Setting $x=y=\frac{a}{2},$ we get $$f(a)=f(0)=1$$for all $a.$ Note that this function works, and in particular, $f(2017)=1.$

Therefore, the possible values of $f(2017)$ are $\boxed{0,1}.$