Question

Sa se arate ca numarul E=n²m+nm²+n²+m²+2mn+n+m este divizibil cu 2,oricare ar fi numerele m,n ∈N

Answer 1

 n²m+nm²+n²+m²+2mn+n+m=(n²+2nm+m²) + (n²m+n) + (nm²+m)=
 =(n+m)² + n(n+m) + m(n+m) =
 =(n+m)(n+m + n + m)=(n+m) * 2 * (n+m) =
 = 2 * (n+m)² $\,\vdots\;2$