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

Question

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 1

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

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

Answer 2

Assuming that the question reads

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

we have

$$ (1-15)^{2n}-1= 1-(-1)-(2n)15+(2n)(2n-1)\frac{15^2}{2!}+\cdots = 15k $$

Answer 3

$3$ does not divide $14$, and so does not divide $14^n$.

By Fermat's little theorem, $14^{2n} = (14^n)^2\equiv 1 \bmod 3$. Thus $3$ divides $14^{2n}-1$.

Answer 4

$14=15-1 ⇒ 14^{2n}-1=(15-1)^{2n}-1==15 m +1 -1=15 m$