Problem
NT-B1-M06-P003 The Equation \(x^2=2y^2\)
#3
★☆☆☆☆ Level 1 of 5
Prove that \(x^2=2y^2\) has no positive integer solutions.
Show that any solution forces both \(x\) and \(y\) to be even.
If \(x^2=2y^2\), then \(x^2\) is even, so \(x=2u\). Then \(4u^2=2y^2\), hence \(y^2=2u^2\), and \(y\) is also even: \(y=2v\). Then \(u^2=2v^2\), a smaller positive solution. Infinite descent is impossible.
Ask the student to name the decreasing parameter explicitly: \(x+y\).