How do i show this :for every prime $p> 3$ and every integer $k\geq1$ ,${p}^{4k}=1\mod3$?

Question

There are many formula which are a multiple of $3$ for example $n^3+2n$ ,I accross this formula " ${p}^{4k}=1\mod3$" after some computations in WA then My question here is:

How do i show this if it is true :for every prime $p> 3$ and every integer $k\geq1$ then ${p}^{4k}=1\mod3$ ?

Note: I tried to use proof by induction but unfourtinately no General formula for primes known !!! Thank you for any help

Answer 1

Hint:

Any prime $p>3$ is congruent to $\pm1\mod3$.

Answer 2

It doesn't matter that $p$ is prime. Just that 3 doesn't divide $p$.

Let $p = 3q \pm 1$. Then $p^m = (3q \pm 1)^m = $ via binomial theorem $ = 3^mq^m + ..... + 3q*(\pm 1)^{m-1} + (\pm 1)^m$.

So $p^m \equiv (\pm 1)^m \mod 3$. If $m$ is even $p^m \equiv 1 \mod 3$. The $4k$ and the requirement that $p$ be prime were overkill.