NT-B1-M06-P024 A Strong Descent with a Constant

Choose a solution with minimal sum. For \(x\ge y\), look at the second root of the quadratic in \(x\).

Assume a solution exists and choose one with minimal sum \(x+y\). Let \(x\ge y\). If \(y=1\), then \(x^2+1=3x+2\), so \(x^2-3x-1=0\), which has no integer root.

Thus \(y\ge2\). View the equation as a quadratic in \(x\): \(x^2-3yx+(y^2-2)=0\). The second root \(x'=3y-x\) is an integer. By Vieta, \(xx'=y^2-2>0\), so \(x'>0\). If \(x=y\), then \(2y^2=3y^2+2\), impossible, hence \(x>y\). Therefore \(x'=\frac{y^2-2}{x}

This is a hard Book 1 problem: the descent uses a second root, not a common divisor.