ALG-B1-M09-P024 First Vieta descent

Assume a minimal solution and view the equation as a quadratic in the larger variable.

Assume a positive solution exists and choose one with minimal sum \(x+y\). Let \(x\ge y\). If \(x=y\), then \(2x^2=5x^2\), impossible. Thus \(x>y\).

Consider \(X^2-5yX+y^2=0\). One root is \(x\), the other by Vieta is \(x'=5y-x\), and the product of the roots is \(y^2\). Hence \(x'=\frac{y^2}{x}\), so \(x'\) is positive and integral.

Since \(x>y\), \(0

A first encounter with the architecture of Vieta descent, without heavy theory.