XVII OBM

The condition $f(x+1019)=f(x)$ allows to make all computations modulo $1019$, which is a prime number. Hence the problem asks the number of multiplicative functions in $\mathbb{Z}/1019\mathbb{Z}$.

Two such functions are $f(x)\equiv 1(\mathrm{mod}1019)$ and $f(x)\equiv 0(\mathrm{mod}1019)$. From now on suppose $f$ is different from these two functions.

Substituting $y=0$, $f(0)=f(x)f(0)\iff f(0)=0$; substituting $y=1$, $f(x)=f(x)f(1)\iff f(1)=1$. By Euler-Fermat's theorem, $x^{\varphi (1019)}\equiv 1(\mathrm{mod}1019)$ so $f(x{)}^{\varphi (1019)}=f(x^{\varphi (1019)})=f(1)=1\iff f(x)=\pm 1$.

Now let $g$ be a primitive root of $1019$ (that is, $g$ such that the least positive integer $d$ such that $g^d\equiv 1(\mathrm{mod}1019)$ is $\varphi (1019)$). So every $x$ in $\mathbb{Z}/1019\mathbb{Z}$ can be written as $g^k$ for some $k$. Then $f(x)=(f(g){)}^k$. If $f(g)=1$ then $f(x)=0$ if $1019$ divides $x$ and $1$ otherwise; if $f(g)=-1$ then $f(x)=0$ if $1019$ divides $x$, $f(x)=1$ if $x$ is a quadratic residue modulo $1019$ (such residues are $1,g^2,g^4,\dots ,g^{1016}$) and $f(x)=-1$ otherwise.