Problem
ALG-B1-M10-P023 Descent
#23
★★★★★ Level 5 of 5
Prove that \(x^2+y^2=6xy\) has no positive integer solutions.
Choose a solution with minimal sum and view the equation as a quadratic in the larger variable.
Assume a positive solution exists and choose one with minimal \(x+y\). Let \(x>y\). As a quadratic in \(x\), the equation is \(X^2-6yX+y^2=0\). The other root \(x'=6y-x=\frac{y^2}{x}\) is positive, integral, and smaller than \(y\). Then \((x',y)\) is a smaller positive solution, contradiction.
A strong problem on recognizing Vieta descent.