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All-Soviet-Union Mathematical Olympiad

Algebra Difficulty 5.1 AIME, harder Prove it Soviet Union

Problem:
Given the equation $x^{3} + a x^{2} + b x + c = 0$ , the first player gives one of $a$ , $b$ , $c$ an integral value. Then the second player gives one of the remaining coefficients an integral value, and finally the first player gives the remaining coefficient an integral value. The first player's objective is to ensure that the equation has three integral roots (not necessarily distinct). The second player's objective is to prevent this. Who wins?

Solution

Solution:
The first player wins.

The first player starts by choosing $c = 0$ . Now if the second player selects $a$ , then the first player can take $b = a - 1$ . Then the polynomial factorizes as: $x(x + 1)(x + a - 1)$ with integral roots $0$ , $-1$ , $1 - a$ .

If the second player selects $b$ , then the first player can take $a = b + 1$ . Then the polynomial factorizes as $x(x + 1)(x + b)$ with integral roots $0$ , $-1$ , $-b$ .

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