[tex]\displaystyle Notam~a=(1+\sqrt{3})^{2018}~si~b=(1-\sqrt{3})^{2018}. \\ \\ Folosind~binomul~lui~Newton,~observam~ca~a~este~de~forma \\ \\ a=p+q \sqrt{3},~iar~b~este~de~forma~b=p-q \sqrt{3},~unde~p,q \in \mathbb{Z}. \\ \\ Atunci~a+b=2p \in \mathbb{Z}. \\ \\ Observam~ca~1- \sqrt{3} \in (-1,0),~si~deci~b \in (0,1). \\ \\ Atunci~a=2p-b \in (2p-1,2p). \\ \\ Deci~[a]=2p-1. \\ \\ Rezulta~\{a\}=a-[a]=a-2p+1=a-a-b+1=1-b= \\ \\ =1-(1- \sqrt{3})^{2018}.[/tex]
