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Multiple Of Six Difference

Problem

Given that $x$ and $y$ are integer, prove that the difference between the expressions $x^3 + y$ and $x + y^3$ is a multiple of six.


Solution

Without loss of generality suppose that $x^3 + y \ge x + y^3$.

$\begin{align}\therefore (x^3 + y) - (x + y^3) &= x^3 - x - y^3 + y\\&= x(x^2 - 1) - y(y^2 - 1)\\&= x(x - 1)(x + 1) - y(y - 1)(y + 1)\end{align}$

As $x - 1$, $x$, and $x + 1$ are three consecutive integers, at least one of the terms must be even and exactly one of the terms must be a multiple of three, so the product will be a multiple of six. In the same way $y(y - 1)(y + 1)$ will be a multiple of six.

Hence the difference will always be a multiple of six.

Problem ID: 330 (13 Jul 2007)     Difficulty: 3 Star

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