Let $f: \mathbb{R}\rightarrow \mathbb{R}$ be a function satisfying $f(x)f(y)=f(x-y)$. Find all possible values of $f(2017)$.  Enter all the possible values, separated by commas.
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 $\boxed{0,1}.$