Prove that : $10^{5n+2}+(-1)^{n}\cdot 4 \equiv 0 \pmod {13}$ Prove that :
$$10^{5n+2}+(-1)^{n}\cdot 4 \equiv 0 \pmod {13}$$
I don't have enough skills in modular to do it 
Please help
$$10^2\equiv-4\pmod{13}$$

So, the problem reduces to $$10^{5n}-(-1)^n\equiv0\pmod{13}$$

$$\iff10^{5n}\equiv(-1)^n\iff(-10)^{5n}\equiv1$$

Now $(-10)^3\equiv1\implies$ord$_{13}(-10)=3$

So, $3$ must divide $5n\iff 3\mid n$ as $(5,3)=1$
$$10^2\equiv-4\pmod{13}$$

So, the problem reduces to $$10^{5n}-(-1)^n\equiv0\pmod{13}$$

$$\iff10^{5n}\equiv(-1)^n\iff(-10)^{5n}\equiv1$$

Now $(-10)^3\equiv1\implies$ord$_{13}(-10)=3$

So, $3$ must divide $5n\iff 3\mid n$ as $(5,3)=1$