NT-B1-M06-P021 The Equation \(x^2+y^2=3xy\)

Assume \(x>y\) and view the equation as a quadratic in \(x\).

Assume a solution exists and choose one with minimal \(x+y\). If \(x=y\), then \(2x^2=3x^2\), impossible. Let \(x>y\). The equation \(x^2-3xy+y^2=0\), viewed as a quadratic in \(x\), has second root \(x'=3y-x\). By Vieta, \(xx'=y^2\), so \(x'>0\), and since \(x>y\), \(x'=\frac{y^2}{x}

This is a preview of Vieta jumping: not a full heavy block, but the architecture is visible.