$24~$ divides $~p^2 -q^2$, where $~p~$ and $~q~$ are primes greater than $~5$

Question

I know this question is answered, here

httpsss://math.stackexchange.com/questions/507451/suppose-that-p-%E2%89%A5-q-%E2%89%A5-5-are-both-prime-numbers-prove-that-24-divides-p2

But when I was doing the question ,I thought that "Can I make difference of square of $2$ primes a product which is divisible by $4!$ "

While doing it, some examples were just trying to show me something but I don't know what the examples are shown in the pic.

enter image description here

Duplicate of For any prime $,p > 3$, why is $p^2-1$ always divisible by 24? — Bill Dubuque, Jul 02 '19 at 15:16
Note $, p^2-q^2 = (p^2-1)-(q^2-1)$ so it suffices to show that $,24\mid p^2-1,$ for $p$ coprime to $2,3$. There are hundreds of answers here that do so, e.g. see the linked dupe. You need to say why these common proofs don't work for you in order for your question not to be a dupe. — Bill Dubuque, Jul 02 '19 at 15:18

score 3 · Accepted Answer · answered Jul 02 '19 at 11:27

3

The square of an odd number is always $\equiv 1\pmod 8$, and the square of an integer that is not a multiple of $3$ is always $\equiv 1\pmod 3$. Hence the difference of two squares with both these properties is a multiple of both $8$ and $3$.

answered Jul 02 '19 at 11:27

Hagen von Eitzen

888

$24~$ divides $~p^2 -q^2$, where $~p~$ and $~q~$ are primes greater than $~5$

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