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Algebra Difficulty 5.0 AIME Prove it India

For a positive integer $n$ , a cubic polynomial $p(x)$ is said to be $n$ -good if there exist $n$ distinct integers $a_1, a_2, \dots, a_n$ such that all the roots of the polynomial $p(x) + a_i = 0$ are integers for $1 \le i \le n$ . Given a positive integer $n$ prove that there exists an $n$ -good cubic polynomial.

Solution

Let $f(x) = x^3 - mx^2 + nx$ , $k$ an integer, and $a, b, c$ the roots of $f(x) + k = 0$ . Then
$4m^2 - 12n = 3(a-c)^2 + (a - 2b + c)^2.$
Since the equation $x^2 + 3y^2 = 1$ has a rational solution, it has infinitely many rational solutions. Therefore one can find $D$ such that $4D$ can be expressed as $3s^2 + t^2$ in at least $6n$ ways. By choosing $m$ and $n$ such that $4m^2 - 12n = 4D$ we get an $n$ -good cubic polynomial. $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.