$(p-1)(p+1)/24 \in \mathbb N$ for all primes $p \geq 5$ I want to show 
\begin{align}
\frac{(p-1)(p+1)}{24} \in \mathbb N \quad \text{for all primes} \quad p \geq 5 \tag{1}.
\end{align}
I can show $(1)$, if the following statement is true.

Let $a,b,c,d \in \mathbb N$ and $a \geq b \cdot c \cdot d$.
  \begin{align}
\text{If} \quad  \frac{a}{b},\frac{a}{c},\frac{a}{d} \in \mathbb N, \quad \text{then} \quad \frac{a}{b \cdot c \cdot d} \in \mathbb N \tag{2}.
\end{align}

Given $(2)$ we show that $(1)$ is true for $a = (p-1)(p+1)$, $b = 2$, $c = 3$ and $d = 4$. Since $p$ is a prime $(p-1)$ and $(p + 1)$ are even, implying $(p-1)/2 \in \mathbb N$, $(p+1)/2 \in \mathbb N$ and thus $(p-1)(p+1)/2 \in \mathbb N$ and $(p-1)(p+1)/4 \in \mathbb N$. One of the three numbers $(p-1)$, $p$ and $(p+1)$ must be divisible by 3. Since $p$ is a prime either $(p-1)$ or $(p+1)$ is divisible by 3, implying $(p-1)(p+1)/3 \in \mathbb N$.
Question Is $(2)$ true?

There are two cases : Either $p = 4k+1$ or $4k+3$ ,
Case 1 :
$$ \begin {align}p &=4k+1 \\
p^2-1 &= 16k^2 +1+8k-1 = \color{red}{8k\,(2k+1)} \end{align}$$
which is clearly divisible by $8$.
Now $K $can be of two form $k = 3m \,, 3m+1$.If $k = 3m+2$ , then $p$ is not prime.
If $k = 3m$ , it is clear that $p^2-1$ is divisible by $24$.
If $k = 3m+1$ , then $(2k+1)$ is divisible by $3$
Case 2 :
$$\begin{align} p &=4k+3 \\
p^2-1 &= 16k^2 +9+24k-1 = 8(2k^2+3k+1) = \color{blue}{8\,(k+1)\,(2k+1)} \end{align}$$
which is clearly divisible by $8$.
Now $K $can be of two form $k =3m+1 ,3m+2 $.
If $k = 3m+1$ , it is clear that $2k+1$ is divisible by $3$.
If $k = 3m+2$ , then $(k+1)$ is divisible by $3$
Which completes all the cases and proves your result.
