elementary number theory - Prove that $3$ divides $14^{2n}-1$ with modular arithmetic.

Like the title says. I know how I would go about solving this with a proof by induction method, but since they ask me to do it with modular arithmetic I'm lost.

I changed it to:

$14^{2n}-1 \cong 0\pmod 3$

And tried to solve it in some way but failed.

Edit: I forgot the -1

Answer

I'm going to follow the trend and assume you want $14^{2n}-1$ to be divisible by $3$. Then we know

$$ 14 \equiv 2 \equiv -1 \mod 3$$

(To check this, just divide $14$ by $3$ and look at the remainder.)

Thus

$$14^{2n}-1 \equiv (-1)^{2n}-1 \equiv ((-1)^2)^n-1 \equiv 1^n-1 \equiv 1-1 \equiv 0 \mod 3$$