Problem
NT-B2-M08-P018 A Mixed Quadratic Form
#18
★★★★☆ Level 4 of 5
Prove that \(x^2+2y^2=5z^2\) has no nonzero integer solutions.
Consider squares modulo \(5\).
Squares modulo \(5\) are \(0,1,4\). Thus \(2y^2\) has residues \(0,2,3\). The sum \(x^2+2y^2\) can be \(0\pmod 5\) only if both terms are \(0\). Hence \(5\mid x,y\). Then \(25\mid x^2+2y^2=5z^2\), so \(5\mid z\). Dividing all variables by \(5\), we get a smaller solution of the same equation. Contradiction.
The key point is that the form is preserved after division by \(5\).