
Prove
Prove that is p (greater than or equal to) 5 is a prime then (p^2) is congruent to 1(mod24).
I have a hint that a positive integer k divides on of k sequential integers n,(n+1),(n+2),...,(n+k1), where n is an integer. the product of any k sequential integers n(n+1)(n+2)...(n+k1) is congruent to 0(mod k)

$\displaystyle p^21=(p1)\cdot{(p+1)}\equiv{0(\bmod.24)}$
We will show that $\displaystyle (p1)\cdot{(p+1)}$ is multiple of 3 and 8, which is equivalent to show that $\displaystyle p^2\equiv{1}(\bmod.24)$
Indeed, since $\displaystyle p$ is prime and $\displaystyle p>3$ we have $\displaystyle p\equiv{1;2}(\bmod.3)$ thus $\displaystyle p^2\equiv{1}(\bmod.3)$ and $\displaystyle (p1)\cdot{(p+1)}$ is multiple of 3
$\displaystyle p$ is prime and $\displaystyle p>2$ so $\displaystyle p\equiv{1;3;5;7}(\bmod.8)$ thus $\displaystyle p^2\equiv{1}(\bmod.8)$ (try each case)
Therefore $\displaystyle (p1)\cdot{(p+1)}$ is multiple of 3 and 8