Problem
NT-B2-M02-P016 Descent for the Number \(3\)
#16
★★★★☆ Level 4 of 5
Prove that the equation \(x^2+y^2=3z^2\) has only the zero solution in integers.
Modulo \(3\), if a sum of two squares is divisible by \(3\), both squares are divisible by \(3\).
Squares modulo \(3\) are \(0\) or \(1\). If \(3\mid x^2+y^2\), the sum of residues cannot be \(1+0\) or \(1+1\), so both squares have residue \(0\). Thus \(3\mid x\) and \(3\mid y\). Then \(9\mid x^2+y^2=3z^2\), so \(3\mid z\). Dividing all variables by \(3\) gives a smaller solution. A nonzero solution would lead to infinite descent, so only \(x=y=z=0\) is possible.
This is the first descent problem in Number Theory Book 2.