Problem
NT-B2-M08-P009 The Second-Root Lemma
#9
★★★☆☆ Level 3 of 5
Let positive integers \(x,y\) satisfy \(x^2+y^2+c=mxy\), where \(c,m\) are integers. Prove that if \(y'=mx-y\) is positive, then \((x,y')\) is also a solution.
View the equation as a quadratic in \(y\).
Write \(Y^2-mxY+x^2+c=0\). One root is \(y\). By Vieta's formula, the sum of the roots is \(mx\), so the second root is \(y'=mx-y\). Since the coefficients are integers and \(y'\) is assumed to be a positive integer, substituting the second root again makes the quadratic zero. Hence \(x^2+(y')^2+c=mxy'\).
This lemma makes Vieta jumping conceptual rather than mechanical.