## Solution.

An integer root of a polynomial with integer coefficients must be a divisor of the free term. Since, without loss of generality, we can assume that the free term of the polynomial Lucija chooses is different from zero, we conclude that the set $S$ can only be extended by integer divisors of the number 95. Since the number of divisors of the number 95 is finite, Lucija will not be able to extend the set $S$ infinitely.

Let's prove that Lucija will add all integer divisors of 95 to the set $S$, i.e., that $S$ will eventually have 9 elements.

Since -1 is a root of the polynomial $95 x+95$, the number -1 can be added to $S$.

The number 1 is a root of the polynomial $-x^{95}-x^{94}-\cdots-x+95$, so 1 can be added to $S$.

Now the number -95 can be added because it is a root of the polynomial $x+95$.

The polynomial $-x^{3}+x^{2}+x+95$ with coefficients from $S$ has a root 5, so 5 can be added to $S$. Now we see that -5 can also be added to $S$ because -5 is a root of the polynomial $x+5$.

Similarly, at the end, $-19,19 \in S$ because they are roots of the polynomials $5 x+95$ and $5 x-95$.

Note: The polynomial for which 5 is a root can be obtained in a systematic way. Clearly, 5 cannot be obtained as a root of a linear polynomial with coefficients from the set $S=\{-95,-1,0,1,95\}$.

Assume that 5 is a root of the polynomial $a x^{2}+b x+c$, for $a, b, c \in S$. Then $5 \mid c$ and $c \neq 0$, so $c=95$ or $c=-95$. We have

$$25 a+5 b= \pm 95$$

or

$$5 a+b= \pm 19$$

which is impossible for $a, b \in S$. Therefore, if 5 is a root of a polynomial with coefficients from $S$, the polynomial must be of degree at least 3.

Assume that

$$5^{3} a+5^{2} b+5 c+d=0$$

for some $a, b, c, d \in S$. Then $5 \mid d$ and $d \neq 0$, so $d=95$ or $d=-95$.

If $d=95$, dividing the above equation by 5 gives

$$25 a+5 b+c=-19$$

so we see that $c \equiv 1(\bmod 5)$. Since $c \in S$, it must be $c=1$.

Now $5 a+b=-4$ from which it easily follows that $a=-1$ and $b=1$.