[tex]\sqrt{(n^{2}+n+1)(n^{2}+n+3)+1}\\Notam:\\n^{2}+n+1=a\\
a(a+2)+1=a^{2}+2a+1=(a^{2}+a)+(a+1)=a(a+1)+1(a+1)\\=(a+1)(a+1)=(a+1)^{2}\\\sqrt{(a+1)^{2}}=a+1=n^{2}+n+2\\
Cum\ n\ este\ natural\ n^{2}=natural\ n=natural\ 2=natural\\
Suma\ a\ trei\ numere\ naturale\ va\ da\ un\ numar\ natural[/tex]
