Problem
NT-B2-M08-P010 A Parametric Impossibility
#10
★★★★★ Level 5 of 5
Let \(k\ge 4\) be an integer. Prove that \(x^2+y^2+1=kxy\) has no solutions in positive integers.
Repeat the Vieta jump and check \(x=1\) separately.
Let \(x\le y\) and take a solution with minimal sum. For \(x=1\), we get \(y^2-ky+2=0\). Its discriminant \(k^2-8\) is not a square: if \(k^2-8=t^2\), then \((k-t)(k+t)=8\), impossible for \(k\ge 4\). Now let \(x\ge 2\). The second root is \(y'=kx-y=\frac{x^2+1}{y}\), positive and integral. Since \(y\ge x\), \(y'\le x+\frac{1}{x}
A stronger version of Problem 6; a good first Level 5 problem in the module.