Problem
NT-B2-M08-P005 No Primitive Solution
#5
★★★☆☆ Level 3 of 5
Prove that \(x^2+y^2=4z^2\) has no solutions with \(\gcd(x,y,z)=1\).
Compare modulo \(4\).
The right side is divisible by \(4\). A square modulo \(4\) is \(0\) or \(1\). For \(x^2+y^2\equiv 0\pmod 4\), both squares must be \(0\), so \(x,y\) are even. Thus \(x,y\) have common divisor \(2\), so the solution cannot be primitive.
A gentle preparation for extracting a common divisor.