Prove that : $10^{5n+2}+(-1)^{n}\cdot 4 \equiv 0 \pmod {13}$

Question

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

Answer 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 $(5,3)=1$

Answer 2

Since $(-3)^{5n+2}=(-1)^n3^{5n+2}$, we have $$(-1)^n(3^{5n+2}+4)$$ on left hand side. Since $3^2 \equiv -4 \mod 13$ we can write as $$\equiv (-1)^n4(1-3^{5n})$$ and this is equivalent to $$(-1)^n4(1+9^{n})$$ or $$(-1)^n4(1+(-4)^{n})$$