\(x^2+y^2=z^2\)
Ta có: \(x^2+y^2-z^2-\left(x+y-z\right)=x\left(x-1\right)+y\left(y-1\right)-z\left(z-1\right)⋮2\)
nên \(\left(x^2+y^2-z^2\right)\equiv\left(x+y-z\right)\left(mod2\right)\)
suy ra \(x+y-z⋮2\Leftrightarrow x-y+3z⋮2\).
Mà \(x+3z-y>x+2z>2\)
Do đó \(x+3z-y\)là hợp số.