# Prove that p^2 – 1 is Divisible by 24

Problem:
Prove that p2 – 1 is divisible by 24 if p is a prime number greater than 3?

Solution:
The most elementary proof , without explicitly mentioning any number theory: out of the three consecutive numbers p–1, p, p+1, one of them must be divisible by 3; also, since the neighbors of p are consecutive even numbers, one of them must be divisible by 2 and the other by 4, so their product is divisible by 3⋅2⋅4=24 — and of course, we can throw p out since it’s prime, and those factors cannot come from it.

### 0 Thoughts on “Prove that p^2 – 1 is Divisible by 24”

1. Suneet Agrawal on April 29, 2015 at 5:27 pm said:

on what basis, you took consecutive numbers p–1, p, p+1.
Is it through the relation:
(A)^2 – (B)^2 = (A+B) (A-B)

if this is the logic, it should be p-1, p+1
how come p ?

• it is (p-1) and (P+1)… P is prime no. >3 and it can’t be divided by 3… either P-1 or P+1 will always be divisible by 24..

2. Vinay on June 18, 2015 at 1:57 pm said:

P^2 – 1 = 24
p^2 = 24 + 1 = 25
p = +/- 5.

Is’nt it enough to prove, p is a prime Number in itself.
Why We need to think complex ways for obvious things.

• Germán on August 28, 2015 at 2:41 am said:

You are assuming that P^2 – 1 = 24, but the problem says P^2 – 1 is divisible by 24.
So, is not as simple as you think.

3. kamal on July 5, 2015 at 5:33 pm said:

according to this question

p^2 – 1 = 24*X let X>0

after solving this equation we will get

p = sqrtof(24*X+1)

now put the values of X

when X=1 , p=5
when X=2 , p=7
when X=5 , p=11

we will not consider X=3 and X=4 because we will not get the perfect squares…..:)

4. Shashikant Deshpande on July 25, 2015 at 3:14 pm said:

p^-1 = (p-1)(p+1)

p is prime, so odd, not divisible by 3 (ignoring 2 & 3).
Therefore, p-1 and p+1 are 2 consecutive even numbers, divisible by 2 & 4, and one of them divisible by 3.
The product therefore is a multiple of 2x3x4 i. e. 24.

5. Alicia Sharma on August 1, 2015 at 3:30 pm said:

6. All prime nos >3 will be in form either 6n+1 or 6n-1

Squaring gives 36n^2+12n+1 or 36n^2-12n+1

subtracting 1 gives 12n(3n+1) or 12n(3n-1)

Now if n is even, it will be multiple of 24 and if n is odd then 3n+1 or 3n-1 will be even hence it will be multiple of 24

7. I think p-1 or p+1 either one of them should be divisible by 6 & 4 so no matter whatever p is the product of p+1 & p-1 will be divisible by 24

8. P^2-1/ 25 , p>3 PRIME then consider prime numbers from 5 replace 5 with p (eliminating process) then the value of P^2 -1 will be a multiple of 24

9. Jitesh Mittal on September 12, 2016 at 4:53 pm said:

A prime no > 3 can be expressed as (6k +1) or (6k-1)

so p^2 – 1= (6k+1)^2 – 1= 36k^2 + 12k = 12k(3k+1)
for 6k-1,
p^2 – 1= (6k-1)^2 – 1= 36k^2 – 12k = 12k(3k-1)

So in both the cases 3k + 1 or 3k – 1 , both multiple of 2. so P ^2 – 1 is multiple of 24.

Thanks