Problem
NT-B1-M11-P022 Equation with No Solutions
#22
★★★★☆ Level 4 of 5
Prove that \(x^2+y^2=3xy\) has no positive integer solutions.
Try a minimal solution and the second root of a quadratic equation.
Assume a solution exists and choose one with minimal \(x+y\). Let \(x\ge y\). View the equation as quadratic in \(x\): \(x^2-3yx+y^2=0\). The second root \(x'=3y-x\) is integral. Since \(x<3y\), it is positive. Also \(x(3y-x)=y^2\); for \(y\le x\le2y\), the left side is at least \(2y^2\), impossible, so \(x>2y\). Hence \(0
Strong introduction to descent.