Problem
NT-B1-M06-P013 The Equation \(x^2+2y^2=5z^2\)
#13
★★★☆☆ Level 3 of 5
Prove that \(x^2+2y^2=5z^2\) has no positive integer solutions.
First apply the result about divisibility of \(x^2+2y^2\) by \(5\).
From \(5\mid x^2+2y^2\), we get \(5\mid x\), \(5\mid y\). Let \(x=5x_1\), \(y=5y_1\). Then \(25x_1^2+50y_1^2=5z^2\), so \(z^2=5x_1^2+10y_1^2\), hence \(5\mid z\). Dividing all variables by \(5\) gives a smaller positive solution. Descent is impossible.
This already has two steps: a modular lemma and the descent itself.