Question

pliss dau coroana .Aratati ca ,oricare ar fi n E N ,numarul a=n³-n este divizibil atât cu 2 cat si cu 3.​

Answer 1

$\displaystyle\it\\n^3-n=n(n^2-1)=n(n^2-1^2)=(n-1)n(n+1).\\numarul~evident~este~divizibil~cu~2~pentru~ca~in\\produs~avem~cel~putin~doua~numere~naturale\\consecutive.\\orice~numar~natural~poate~fi~de~forma~\mathcal{M}_3,\mathcal{M}_3+1,\mathcal{M}_3+2.\\fie~k\in\mathbb{N}\\daca~n=3k \implies evident~a~este~divizibil~cu~3~si~cu~2.\\daca~n=3k+1 \implies (n-1)=3k \implies evident~a~este~divizibil~cu\\3~si~cu~2.\\daca~n=3k+2 \implies(n+1)=3k+2+1=3(k+1)\implies\\a~este~divizibil~atat~cu~3~cat~si~cu~2.$