Problem
NT-B1-M03-P006 Sum of Two Squares and \(4z+3\)
#6
★★☆☆☆ Level 2 of 5
Prove that \(x^2+y^2=4z+3\) has no integer solutions.
Consider the equation modulo \(4\).
A square modulo \(4\) is \(0\) or \(1\). Hence \(x^2+y^2\) can have residue \(0,1,2\), but not \(3\). The right side has residue \(3\). Contradiction.
Standard impossibility via modulo \(4\).