Olympiad Maths Prep

Library / /17 of 60

Algebra Difficulty 5.3 AIME, harder Prove it Ukraine

Find possible positive solutions (x,y,z)(x, y, z) for the given system of equations:
{x+y2=2z3,y+z2=2x3,z+x2=2y3. \begin{cases} x + y^2 = 2z^3, \\ y + z^2 = 2x^3, \\ z + x^2 = 2y^3. \end{cases}

Solution

Without loss of generality let z≥max⁡{x,y}z \ge \max\{x, y\}. Consider the following cases.

Case 1. z>1z > 1. Then due to the first equation x+y2≤z+z2<2z3x + y^2 \le z + z^2 < 2z^3 – contradiction.

Case 2. z=1z = 1. Due to the first equation, 2=x+y22 = x + y^2.
Since max⁡{x,y}≤1\max\{x, y\} \le 1, it is possible only if x=y=1x = y = 1. It is easy to check that (1;1;1)(1; 1; 1) is a solution.

Case 3. z<1z < 1. Then for the smallest variable, say xx, x<1x < 1 holds, then by the second equation:
x+y2≤z+z2<2z3x + y^2 \le z + z^2 < 2z^3 – contradiction.

Similar contradiction holds if yy is the smallest variable.

Looking for a route rather than an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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