Problem
NT-B1-M06-P018 The Equation \(x^2=45y^2\)
#18
★★★☆☆ Level 3 of 5
Prove that \(x^2=45y^2\) has no positive integer solutions.
First use divisibility by \(3\), then reduce to \(\sqrt{5}\).
Since \(3\mid x^2\), we have \(x=3u\). Then \(9u^2=45y^2\), hence \(u^2=5y^2\). But \(u^2=5y^2\) has no positive integer solutions: it forces both \(u\) and \(y\) to be divisible by \(5\), starting descent. Contradiction.
Shows how to handle a coefficient containing a square factor.