Problem
NT-B1-M06-P022 Discriminant and Descent
#22
★★★★☆ Level 4 of 5
Prove that \(x^2+y^2=4xy\) has no positive integer solutions.
View the equation as a quadratic in \(x\).
As a quadratic in \(x\), the equation is \(x^2-4yx+y^2=0\). Its discriminant must be a square: \(D=16y^2-4y^2=12y^2\). Thus there is an integer \(u\) such that \(u^2=12y^2\). But this equation has no positive integer solutions, as proved earlier. Contradiction.
Connects Diophantine equations with irrationality of roots.