Proving modular arithmatic congruence

Question

Let $p \in \mathbb{N}$ be prime with $p \gt 3$. Prove that $p^2 \equiv 1 \pmod {24}$.

Ideas I have so far:

• A prime must be odd, and an odd squared is odd
• Use induction?
• I know $24$ divides $1 - p^2$
• $1 + 24a = p^2$ where $p$ is a prime greater than $3$ and $a$ is an integer

I believe I have everything that I need to finish this problem, I just dont quite know how to put them together. Can anyone give me a tip/hint on how to finish? (An answer is not what I'm looking for :)) Thanks!

Answer 1

Observe that $\;p>3\;$ a prime$\;\implies\;p=\pm1\pmod 6\iff p=\pm1+6k\;,\;\;k\ge1\;$ , so

$$p^2=36k^2\pm12k+1\implies p^2-1=12k(3k\pm1)\;(**)$$

and observe now that $\;k\;$ odd $\;\implies 3k\pm1\;$ even $\;\implies\;(**)\;$ is divisible by $\;2^3\;$ and $\;3\;$ , and the same if $\;k\;$ is even

Answer 2

$p^2-1=(p-1)(p+1)$

Since $p>3$:

one of $p-1$, $p+1$ is divisible by 3,

both $p-1$, $p+1$ are even, one of them is divisible by 4.

So $p^2-1=(p-1)(p+1)$ is divisible by $3\cdot 2\cdot 4=24$

Answer 3

$p^2-1 = (p-1)(p+1)$

p is odd,
p+1 is even, p-1 is even, and one of the two must be divisible by 4.

p is not divisible by 3, so one of p+1 or p-1 is divisible by 3.