inceraca a=3*n+1 sau a=3*n+2, atunci a la patrat = 3*m+1
Caz1) n=3*c+1 => n^2=(3*c+1)(3*c+1) => n^2=9*c^2+6*c+1 => n^2=3*(3*c^2+2*c)+1
Caz2) n=3*c+2 => n^2=(3*c+2)(3*c+2) => n^2=9*c^2+12*c+4 => n^2=3*(3*c^2+4*c+1)+1
N1^2=3*C1+1
N2^2=3*C2+1
N2013^2=3*C2013+1
obtinem ca N1^2+N2^2+ ... +N2013^2=3*(C1+C2+ ... +C2013)+2013*1=3*(C1+C2+ ... +C2013+671).
Asadar, suma patratelor a 2013 nr. naturale nedivizibile cu 3 este divizibila cu 3.
