Show that if $p$ is a prime number $> 3$ then $24 \mid p^2-1$ Hi guys can someone help me with this ?(Without using Modular arithmetic)
Show that if $p$ is a prime number $>3$ then $24$ $\mid$ $p^2-1$
 A: Without (explicitly) using modular arithmetic:
$p^2-1=(p+1)(p-1)$. Since $p$ is odd, this is a product of two even numbers that differ by two, so one of them is in fact divisible by four, and their product is divisible by 8.
Now, $p-1$, $p$ and $p+1$ are three consecutive numbers, so 3 divides one of them. $p$ is prime, so it's not divisible by three. Thus 3 certainly divides $p^2-1=(p-1)(p+1)$.
We conclude that since both 3 and 8 divide $p^2-1$, so too does 24 (using the fact that 3 and 8 are coprime).
A: If $p$ is a prime greater than $3$, than it is equal to $1, 3, 5, 7$ $\pmod 8$
THis implies $p^2 \equiv 1 \pmod 8$ (because $1^2 \equiv 1$, $3^2 \equiv 1, 5^2 \equiv 1, 7^2 \equiv 1 \pmod 8$)
Also, $p \equiv 1, 2 \pmod 3 \Rightarrow p^2 \equiv 1 \pmod 3$
THis implies that $p^2 - 1 $ is divisible by both $8$ and $3$, and so it is divisible by $24$ 
A: If the prime is greater than 3 then it is either of the form $3m+1$ or $3m+2$ and in both the cases $p^2-1$ is divisible by 3. Primes above three are odd. So on squaring them they leave remainder 1 when divided by 8. Hence $p^2-1$ is divisible by 8. So $p^2-1$ is divisible by 24.
A: We know that any prime number $>3$ can be written as $6n\pm1$. Ergo $$p^2-1=(6n\pm1)^2-1=36n^2\pm12n+1-1=36n^2\pm12n=12n(3n\pm1)$$ Now note that one between $n$ and $3n\pm1$ must be divisible by $2 \Rightarrow$ $$p^2-1 = 12n(3n\pm1) \text{ is divisble by $12\cdot 2=24$}$$
