Prove that p^2 – 1 is Divisible by 24

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

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 ?

  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.

  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:

    Nice puzzle. Great application of your brain. Enhances your mathematical skills.

  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.


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