Problem
NT-B2-M08-P006 No Solution With Coefficient \(4\)
#6
★★★★☆ Level 4 of 5
Prove that \(x^2+y^2+1=4xy\) has no solutions in positive integers.
Choose a solution with minimal sum and view the equation as a quadratic in the larger variable.
Let \(x\le y\) and take a minimal solution. As a quadratic in \(y\), the equation is \(Y^2-4xY+x^2+1=0\). The second root is \(y'=4x-y=\frac{x^2+1}{y}\), positive and integral. Since \(y\ge x\), \(y'\le x+\frac{1}{x}\), so \(y'\le x\). Equality \(y'=x\) would give \(y=3x\), hence \(x^2+9x^2+1=12x^2\), or \(2x^2=1\), impossible. We get a smaller solution. For \(x=1\), the discriminant is \(16-8=8\), not a square. Contradiction.
The first full Vieta jump in the problem set.