<s>[INST] $(p-1)(p+1)/24 \in \mathbb N$ 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?
 [/INST] Statement (1) can also be shown by realizing that all primes $p\ge 5$ are odd numbers having no factors of $2$ or $3$, and hence have the form $6k\pm 1$. 
$(p-1)(p+1)=p^2-1=(6k\pm 1)^2-1=36k^2 \pm 12k+1-1=12\cdot k \cdot (3k\pm 1)$
$k$ must be either odd or even. If $k$ is even, then the product has at least one more factor or $2$ in addition to the explicit factor of $12$ and hence is a multiple of $24$.  If $k$ is odd, then $(3k\pm 1)$ is even and the product has at least one more factor or $2$ in addition to the explicit factor of $12$ and hence is a multiple of $24$. 
</s>