[tex]\it n=1986^2-1986-1985=1986^2-1-1985-1985=(1985^2-1^2)-2\cdot1985=\\ \\ =(1986-1)(1986+1)-2\cdot1985=1985\cdot1987-2\cdot1985=1985(1987-2)=\\ \\ =1985\cdot1985[/tex]
[tex]\it \dfrac{\dfrac{x}{1985}}{5}=\dfrac{x}{1985\cdot5}\\ \\ \\ Acum,\ propor\c{\it t}ia\ devine:\\ \\ \\ \dfrac{x}{1985\cdot5}=\dfrac{397^{(397}}{ 1985\cdot1985} \Rightarrow \dfrac{x}{1985\cdot5}= \dfrac{1}{1985\cdot5}\Rightarrow x=1[/tex]
