<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] Primes greater than or equal to $5$ can't be divisible by $3$, so must have the form $3k+1$ or $3k+2$ , so one of $p-1$ or $p+1$ has a factor of $3$. Also, $p-1$ and $p+1$ are consecutive even numbers and one of them is divisible by $4$, so $2$ divides $(p-1)(p+1)$ with at least multiplicity $3.$ 
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