# $(p-1)(p+1)/24 \in \mathbb N$ for all primes $p \geq 5$ [duplicate]

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?

• m.youtube.com/watch?v=ZMkIiFs35HQ – user645636 Nov 15 '19 at 12:03
• Your title question differ from what you are asked in body!! – C.F.G Nov 16 '19 at 4:16

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 $$3.$$

$$(2)$$ 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 $$\quad(p-1)(p+1)\equiv (-4)\cdot (-2)\equiv 0\mod 8$$.

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 $$t$$.

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 $$\text{lcm}(8,3) = 24$$.

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.

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 $$24$$.