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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!

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  • $\begingroup$ @mathlove you're right. thank you, I updated it. $\endgroup$ – S.Lim Feb 25 '16 at 18:33
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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];

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

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    $\begingroup$ Wonderful! thank you. but how did you find that number? just computation? or is there any other systematic way? $\endgroup$ – S.Lim Feb 25 '16 at 18:52
  • $\begingroup$ $0,1,-1$ are bad choices for the boy's next choice $n$, and for any $p$ with $p\mid n$ we better have $p^2\mid n$ (all these observations follow from what happens if the girl makes $C=n$). So trying squares is probably a good idea. A simplification might be to let $a+b=1$ for the $C=n$ case. Then $0=ab+ac+bc$ measn $c=-ab$ and $n=-abc=c^2=b^2(b+1)^2$ (i.e., one of $4,36,144,400,900,1764,\ldots$) $\endgroup$ – Hagen von Eitzen Feb 25 '16 at 19:12

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