$(p-1)(p+1)/24 \in \mathbb N$ 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?
 A: 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 $3.$ 
A: $(2)$ is true only if $\:b,c,d\:$ are pairwise coprime. 
This being said, the simplest way to prove $(1)$ uses congruences and the Chinese remainder theorem:
To prove that $24$ divides $(p-1)(p+1)$ for every prime $p\ge 5$, you just have to show that this product is congruent to $0$ modulo $3$ and modulo $8$.


*

*Modulo $3$, an odd prime $p\equiv \pm1$, so $\;(p-1)(p+1)\equiv 0\cdot 2$ or $-2\cdot 0\mod 3$.

*Modulo $8$, $p\equiv \pm 1$ or $\pm 3$, so
– if $p\equiv\pm 1$, we have $(p-1)(p+1)\equiv 0\mod 8$ for the same reason as above.
– if $p\equiv\pm3$, $\;(p-1)(p+1)\equiv 2\cdot4\equiv 0\;$ or  $\quad(p-1)(p+1)\equiv (-4)\cdot (-2)\equiv 0\mod 8$. 
A: The claim is false. Take $a=6$, $b=1$, $c=2$, $d=2$.
To prove (1), just note that $8$ divides $t^2-1$ for every odd integer $t$.
A: 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 $\text{lcm}(8,3) = 24$.
A: 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.
A: 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 $24$. 
