$(p-1)(p+1)/24 \in \mathbb N$ 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?

As written $(2)$ is not true. Consider $(a,b,c,d) = (36,2,3,4)$. 
Clearly, $36 \ge 24 = 2 \cdot 3 \cdot 4$ and $\dfrac{36}{2} = 18$, $\dfrac{36}{3} = 12$, $\dfrac{36}{4} = 9$ are all positive integers. 
However, $\dfrac{36}{24} = \dfrac{3}{2}$, which is not an integer. 
If you know that $\gcd(b,c) = \gcd(c,d) = \gcd(b,d) = 1$, then $\dfrac{a}{b},\dfrac{a}{c},\dfrac{a}{d} \in \mathbb{N} \implies \dfrac{a}{bcd} \in \mathbb{N}$ is true. 
For the problem you are given, notice that $24 = 2^3 \cdot 3$. So if you can show that $(p-1)(p+1)$ is divisible by $2^3 = 8$ and $(p-1)(p+1)$ is divisible by $3$, then it must hold that $(p-1)(p+1)$ is divisible by $\text{lcm}(8,3) = 24$.
