Problem
NT-B1-M06-P015 The Equation \(x^2+y^2=7z^2\)
#15
★★★☆☆ Level 3 of 5
Prove that \(x^2+y^2=7z^2\) has no positive integer solutions.
Use the previous problem, then divide all variables by \(7\).
The equation implies \(7\mid x^2+y^2\), so \(7\mid x\) and \(7\mid y\). Let \(x=7x_1\), \(y=7y_1\). Then \(49x_1^2+49y_1^2=7z^2\), hence \(z^2=7x_1^2+7y_1^2\), so \(7\mid z\). Dividing all variables by \(7\) gives a smaller solution of the same type. Infinite descent is impossible.
Analogue of the modulo \(3\) problem, but with a richer residue table.