Polynomial game problem: do we have winning strategy for this game?

Question

I'm thinking about some game theory problem. Here it is,

Problem: Consider the polynomial equation $x^3+Ax^2+Bx+C=0$. A priori, $A$,$B$ and $C$ are "undecided", yet and two players "Boy" and "Girl" are playing game in following way:

1. First, the "Boy" chooses a real number.
2. Then, the "Girl" decides the position of that number among $A$,$B$ and $C$.
3. They repeat this process three times, deciding all values for $A$,$B$ and $C$.

If the final polynomial has all distinct integer roots, the Boy wins. Otherwise, the Girl wins.

Question is: Does one of Boy or Girl has any "Winning Strategy"? If so, who has one?

It seems to me, obviously the boy has great advantage. Here is my attempted argument: If the boy suggest "$0$" at the very first turn, regardless of the girl's decision we can make the polynomial to have three distinct integer roots. Actually, my argument has "almost" worked. for example,

1. If the girl put $0$ at position $A$, then boy should suggest "$-84$" at the second round. Then, in any case, we always have three distinct roots, i.e. $(10,-2,-8)$ or $(-3,-4,7)$.
2. If the girl put $0$ at position $C$, then boy should suggest "$-1$" at the second round. Then, in any case, we always have three distinct roots, i.e. $(-1,2,0)$ or $(1,-1,0)$.
3. HOWEVER, if the girl put $0$ at position $B$, I couldn't find any good number for second round.

Has my winning strategy some problem? Or has the girl a winning strategy somehow?

Thank you for any help in advance!

Answer 1

If the girl puts $0$ at position $B$, the boy can choose $1764$, resulting in roots $[-6, 7, 42]$ or $[-864, -1440, 540]$.

EDIT: If the polynomial $(x-a)(x-b)(x-c)$ has $ab+bc+ac=0$, then $c = -ab/(a+b)$. Writing $t = a+b$, we need $t$ to divide $ab = at - a^2$. Thus $t$ is a divisor of $a^2$.

Here's a Maple program that finds the solution $1764$, together with some others. For positive integer values of $a$, it looks at each divisor $t$ of $a^2$ (positive as well as negative), computes the corresponding $b$ and $c$, and the corresponding values of the coefficients $A$ and $C$ of the polynomial, storing under those indices the values $[a,b,c]$. We then look for an index that appears for both $A$ and $C$.

for a from 1 to 1000 do
  for t in numtheory:-divisors(a^2) union map(`*`,numtheory:-divisors(a^2),-1) do
     b:= t-a;
     c:= -a*b/t;
     if nops({a,b,c}) < 3 then next fi;
     A[-a-b-c]:= [a,b,c];
     C[-a*b*c]:= [a,b,c];
  od
od:
{indices(A)} intersect {indices(C)};

$$ \{[-12348], [-1764], [1764], [3136], [12348]\}$$

A[1764], C[1764];

$$[42, 7, -6], [540, -864, -1440]$$