Problem
NT-B1-M06-P005 The Equation \(x^2=3y^2\)
#5
★☆☆☆☆ Level 1 of 5
Prove that \(x^2=3y^2\) has no positive integer solutions.
Use the fact that if \(3\mid x^2\), then \(3\mid x\).
From \(x^2=3y^2\), we get \(3\mid x^2\), so \(x=3u\). Then \(9u^2=3y^2\), hence \(y^2=3u^2\). Therefore \(3\mid y\). Dividing both \(x\) and \(y\) by \(3\) gives a smaller positive solution, impossible by descent.
The same scheme as for \(\sqrt{2}\), but with prime \(3\).