RMO 2005 Questions, Solutions, Discussions

Let \(x, y\) be integers such that \(17\) divides both the expressions \(x^2 - 2xy + y^2 - 5x + 7y\) and \(x^2 - 3xy + 2y^2 + x - y\). Note that \[ x^2 - 3xy + 2y^2 + x - y = (x-y)( x-2y+1), \] which is divisible by \(17\). It follows that \[ x \equiv y \bmod 17, \quad \text{ or } x \equiv 2y -1 \bmod 17 \] holds. Let us consider the case that \(x \equiv y \bmod 17\). It follows that \[ x^2 - 2xy + y^2 - 5x + 7y \equiv (x-y)^2 - 5x+ 7y \equiv 2y \bmod 17. \] Since \(17\) divides \(x^2 - 2xy + y^2 - 5x + 7y\), we get \(2y \equiv 0 \bmod 17\), which yields \(x\equiv y \equiv 0 \bmod 17\), and hence \(17\) divides \(xy - 12x + 15y\). Let us consider the case that \(x \equiv 2y -1 \bmod 17\). Using \(x^2 - 2xy + y^2 - 5x + 7y\equiv 0 \bmod 17\), we obtain \[ (2y -1 )^2 - 2(2y -1 )y + y^2 - 5(2y-1) + 7y\equiv 0 \bmod 17, \] which yields \(y^2-5y + 6 \equiv 0 \bmod 7\). This implies that \((y-2)(y-3)\equiv 0 \bmod 17\). This shows that either \(x\equiv 3\bmod 17, y \equiv 2 \bmod 17\) holds, or \(x \equiv 5 \bmod 17, y \equiv 3 \bmod 17\) holds. If \(x\equiv 3\bmod 17, y \equiv 2 \bmod 17\) holds, then \[ xy - 12x + 15y \equiv 6 - 36 + 30 \equiv 0 \bmod 17 \] holds. If \(x \equiv 5 \bmod 17, y \equiv 3 \bmod 17\) holds, then we obtain \[ xy - 12x + 15y \equiv 15 - 60 + 45 \equiv 0 \bmod 17. \] This proves that \(17\) divides \(xy - 12x + 15y\).

The given inequalities are equivalent to \[ (a-b)^2-c^2 \geq 0 , (b-c)^2 - a^2 \geq 0, (c-a)^2 -b^2 \geq 0 , \] which yields

\[ \begin{align} (a-b+c)(a-b-c) &\geq 0,\ (b-c+a)(b-c-a) &\geq 0,
(c-a+b)(c-a-b) &\geq 0. \end{align} \]

Multiplying them, we obtain \[ -(b+c-a)^2(c+a-b)^2 (a+b-c)^2\geq 0, \] which shows that \((b+c-a)(c+a-b)(a+b-c)\) is equal to \(0\). This proves the result.