[tex]3^{2018}-2^{2018}=(3^2)^{1009}-(2^2)^{1009}=9^{1009}-4^{1009}\\
\text{Mai departe folosim uramtoarea proprietate:}\\
\boxed{a^n-b^n\ \vdots\ a-b,\forall n,a,b\in \mathbb{N}}\\
\text{Asadar:} 9^{1009}-4^{1009}\ \vdots 9-4 ,\text{adica }9^{1009}-4^{1009}\vdots 5[/tex]