If $p > 3$ is prime, then $12 $ divides $p^2 - 1$

Question

First up, I know there are a lot of similar questions with 24, not 12. So bare with me please :)

What is the Question?

Consider the following numbers of the form $p^2 - 1$ where $p$ is prime.

$$5^2 - 1 = 24$$ $$7^2 - 1 = 48$$ $$11^2 - 1 = 120$$

Each of these numbers is divisible by 12. Prove or provide a counter example to the following statement: "If $p > 3$ is prime, then $12 $ | $p^2 - 1$"

What have I tried?

Okay so off the cusp I know two methods that might help.

1. Modular Arithmetic
2. And that any prime $\geq$5 can be represented as $6k \pm 1 $

I'm only just catching up on course content and modular arithmetic so I don't really know where to start.

I assume because 12 | 24, the proof will be very similar to the ones provided on this forum for that question For any prime $p > 3$, why is $p^2-1$ always divisible by 24? I guess I wan't to make sure.

Answer 1

$$(6k\pm1)^2-1=36k^2 \pm12k=12\cdot(3k^2+k)$$

Answer 2

Since u know that any prime $≥5$ can be represented as $6 k±1$ so, $$p=6 k±1$$ $$p^2=36 k^2 ± 12 k + 1$$ $$p^2=12(3 k^2 ± 2 k) + 1$$ $$p^2-1=12(3 k^2 ± 2 k) + 1 -1$$ $$p^2-1=12(3 k^2 ± 2 k)$$ Which is divisible by $12$....(Q.E.D)

Answer 3

Since $p^2-1=(p-1)(p+1)$ and $p$ is odd, you have that $p^2-1$ is the product of two even numbers, hence their product is divisble by $4$. Further, one of $p-1,p,p+1$ is a multiple of $3$, but since $p$ is prime, it is not $p$, hence the product is divisible by 3. Therefore, the result is divisible by $3*4=12$.

Actually, given two consecutive even numbers, one of them will be divisible by $4$, and hence $p^2-1$ will actually be divisible by $8*3=24$.

Answer 4

$\begin{array}{rcl}p^2-1 &=& (p+1)(p-1) \\\\ &=& (6k\pm1+1)(6k\pm1-1) \\\\ &=& (6k+2)(6k)\ \text{or}\ (6k)(6k-2) \\\\ &=& 12k(3k+1)\ \text{or}\ 12k(3k-1)\end{array}$