Croatia_2018

If the coefficients of a polynomial are integers, then its roots must divide the constant term. Without loss of generality, we may assume that the constant term of the chosen polynomial is non-zero, so we conclude that we can extend $S$ only with the divisors of $95$. Since $95$ has only finitely many divisors, Lucija will not be able to extend $S$ indefinitely.

We will prove that Lucija can add all the divisors of $95$ into $S$, i.e. that $S$ will have $9$ elements in the end.

Since $-1$ is the root of the polynomial $95x+95$, we can add $-1$ to $S$.

Number $1$ is a root of $-x^9-x^{94}-\cdots -x+95$, so $S$ can be extended with $1$.

Now we can extend $S$ with $-95$ because it is the root of $x+95$.

Polynomial $-x^3+x^2+x+95$ with coefficients in $S$ has a root $5$, hence $5$ becomes an element of $S$. Now we add $-5$ into $S$, because $-5$ is the root of $x+5$.

In the end, $-19$, $19\in S$ since these are the roots of polynomials $5x+95$ and $5x-95$.