Problem
NT-B2-M08-P011 A Markov Jump
#11
★★★★☆ Level 4 of 5
Let positive integers \(x,y,z\) satisfy \(x^2+y^2+z^2=3xyz\). Prove that \(z'=3xy-z\) is positive and that \((x,y,z')\) is also a solution.
Use Vieta's formulas for the quadratic equation in \(z\).
Consider \(Z^2-3xyZ+x^2+y^2=0\). One root is \(z\), the other is \(z'=3xy-z\). The product of the roots is \(x^2+y^2>0\), so \(zz'>0\); since \(z>0\), \(z'>0\). By Vieta's formula, the second root satisfies the same quadratic equation, so \(x^2+y^2+(z')^2=3xyz'\).
This proves the jump, but not yet the decrease; it is a technical building block.