[tex]\displaystyle\it\\(a+b+c)=par \implies (a+b+c)^2=par,~evident,\\~\Big(2k=par \implies (2k)^2=4k^2=par,~k\in\mathbb{N}\Big)~.\\(a+b+c)^2= a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \Leftrightarrow\\a^2+b^2+c^2=(a+b+c)^2-(2ab+2bc+2ca)=\\\underbrace{(a+b+c)^2}_{par}-\underbrace{2(ab+bc+ca)}_{par}=par,~deci~numarul~nu~poate~fi~Impar.[/tex]
