Răspuns :
[tex] (1+\sqrt5)^2+(1-\sqrt5)^2 = \\ \\ =1^2+2\cdot \sqrt5+\sqrt5^2 + 1^2-2\cdot \sqrt5+\sqrt5^2 = \\ \\ = 1+2\sqrt5+5+1-2\sqrt5+5 = \\ \\ = 1+1+5+5+2\sqrt5-2\sqrt5 = \\ \\ = 12+0 = \\ \\ = 12 \in \mathbb_{N}[/tex]