Problem
NT-B1-M03-P021 The Equation \(x^2+y^2=3z^2\)
#21
★★★★☆ Level 4 of 5
Prove that the only integer solution of \(x^2+y^2=3z^2\) is \(x=y=z=0\).
First prove that if \(x^2+y^2\) is divisible by \(3\), then \(3\mid x\) and \(3\mid y\).
The equation implies \(3\mid x^2+y^2\). By the table of squares modulo \(3\), we get \(3\mid x\) and \(3\mid y\). Let \(x=3x_1\), \(y=3y_1\). Then \(9x_1^2+9y_1^2=3z^2\), so \(z^2=3(x_1^2+y_1^2)\), and \(3\mid z\). Thus all of \(x,y,z\) are divisible by \(3\). If a nonzero solution existed, we could divide all three numbers by \(3\) indefinitely, impossible for nonzero integers. Hence no nonzero solutions exist.
This is a preview of descent; the main descent module comes later.