1. Daca x si y au aceeasi paritate:
x=2m si y=2n, sau x=2m+1 si y=2n+1; m si n ∈Z
x²+y² =(2m) ²+(2n)²=4m²+4n²=4k
x²+y² =(2m+1)²+(2n+1)²= 4m²+4m+1+4n²+4n+1= 4(m²+m+n²+n)+2=4k+2
2. Daca au paritati diferite:
x²+y²=(2m) ²+(2n+1)²=4m²+4n²+4n+1=4k+1
=> x²+y² = {4k; 4k+1; 4k+2}
Deci, x²+y² ≠4k+3; dar, 2023=4•505+3=4k+3
=> ecuatia x²+y²=2023 nu admite solutii in multimea Z
