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.
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..
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.
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.
But it’s still divisible by itself, so why are you saying it’s incorrect?
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…..:)
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.
Nice puzzle. Great application of your brain. Enhances your mathematical skills.
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
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
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
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
what is the value for p
p value is 24
Problem can be expressed as below
P^2 – 1 = 24 . k (for all k > 0)
above expression can be rewritten as
(P + 1) (P – 1) = 24 k (using formula, (a^2 – b^2) = (a + b) (a – b)
Now, we can notice following
Since P is prime,
1. P must be a odd number.
2. P can not be divided by any number since it is prime.
From above 2 observations, we can conclude that
P – 1 must be a even number and P + 1 must be an even number, consecutive numbers in integer set can be expressed as
…., (P – 1), P, (P + 1), ……
Notice,
Since P – 1 is even, P + 1 must be next even number. From this we can conclude that, P + 1 must always divisible by 4. (P – 1) itself is divisible by 2.
Further (P – 1), P, (P + 1) are three consecutive numbers, therefore one of them must be divisible by 3. We also know that P is not divisible by 3. Therefore, (P – 1) or (P + 1) must be divisible by 3.
Putting all these together, we can conclude that
(P – 1) (P + 1) must be divisible by 2, 4, and 3.
In other words,
P^ 2 – 1 must be divisible by 24 which equal to 2 * 4 * 3.