$$2006!+\frac{4012!}{2006!}=x \pmod{4013}$$
Answer: $x=1553.$
Solution: $$2006!+4012!/2006!=x\pmod{4013}$$ $$(2006!)^2 -2006!x+4012!=0\pmod{4013} (*)$$ $$4\cdot (2006!)^2-4\cdot 2006!x+4\cdot 4012!=0$$ $$(2\cdot 2006!-x)^2-x^2+4\cdot 40121=0$$ $$(2\cdot 2006!-x)^2=x^2-4\cdot 4012! \pmod{4013}(1)$$ For $x=1553$ we have $$x^2=1553^2=601\cdot 4013-4 \implies 1553^2=-4\pmod{4013}.$$ On the other hands, Since $4013$ is prime number,by Wilson's theorem we have $(4012!+1);$ $4013$ It follows that $$4012!=-1\pmod{4013} (2)$$ $$4\cdot 4012!=-4\pmod{4013}$$ From $(1),(2),(3)$ we get $$(2\cdot 2006!-1553)^2=1553^2-4\cdot 4012!\pmod{4013} =-4-(-4)=0\pmod{4013}$$ $$(2\cdot 2016!-1553)^2=0\pmod{4013}$$ $$2\cdot 2016!-1553=\pmod{4013}$$ The last equality is always true because of $$2\cdot 2016!=2\cdot (1\cdot 2\cdot 3\cdot \ldots \cdot 2016) 1553$$ Thus,we proved that $(1)$ is true for $x=1553$ or $(*)$ is true for $x=1553.$
Therefore $$2006!+ 4012!/2006!=1553\pmod{4013}*$$
I need a method to solve the example without using the answer, please with full proof.