Competition problem — Number theory

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Number theory Difficulty 5.4 AIME, harder Prove it Romania

a) Prove that, for every integer $k$ , the equation $x^3 - 24x + k = 0$ has at most one integer solution.

b) Prove that the equation $x^3 + 24x - 2016 = 0$ has exactly one integer solution.

Solution

a) Suppose that there exist two different integers $m$ and $n$ so that $m^3 - 24m + k = 0$ and $n^3 - 24n + k = 0$ .
Subtracting the above yields $(m-n)(m^2 + mn + n^2 - 24) = 0$ . Since $m$ and $n$ are different, $m^2 + mn + n^2 = 24$ , whence $(2m+n)^2 + 3n^2 = 96$ .
Therefore $n^2 \le 32$ , hence $n^2 \in \{0, 1, 4, 9, 16, 25\}$ . This leads to $(2m+n)^2 \in \{96, 93, 84, 69, 48, 21\}$ , a contradiction.

b) The equation is $x(x^2 + 24) = 2016$ , so $x$ must be a positive integer; $x = 12$ is a solution.
If $x < y$ are positive solutions, then $x^2 + 24 < y^2 + 24$ and $2016 = x(x^2 + 24) < y(y^2 + 24) = 2016$ , a contradiction.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.