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• 0 Prove or disprove:

If p is a prime number > 3, then p2-1 is divisible by 24.

I love primes Things to note, p is a prime > 3 =>

1. p is odd.

2. p is not divisible by any other number apart from 1 and itself => p is not divisible by 3.

Now the solution:

p2-1 = (p+1)(p-1)

=> p2-1 is product of 2 consecutive even numbers (from 1)

Now product of 2 consecutive even number is always divisible by 8. (lets say 2n*(2n+2) = 4n(n+1)** =4*(2m-1)*(2m-1+1) = 8m(2m-1) )

** - Note: n(n+1) is again a product of consecutive number and hence is divisible by 2 which is shown in the afore mentioned steps

Also note that one of any 3 consecutive number is always divisible by 3, and since it cannot be p (from 2); it has to be either (p-1) or (p+1).

Thus, p2-1 is divisible by 8 and 3. Hence its divisible by 24!

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• 0 First of all, any prime number squared is divisible by 24.

If you are talking about the dividion resulting in an integer, then that's true too.

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• 0 First of all, any prime number squared is divisible by 24.

If you are talking about the dividion resulting in an integer, then that's true too.

No. 51 .. No. The definition of "divisible" given by www.dictionary.com is:

di·vis·i·ble /dɪˈvɪzəbəl/ Pronunciation Key - [di-viz-uh-buhl]

–adjective

2. Mathematics.

a. capable of being evenly divided, without remainder.

b. of or pertaining to a group in which given any element and any integer, there is a second element that when raised to the integer equals the first element.

Also, prime squared is not divisible by any other number apart from that prime number saquared, that prime number and 1.

On the other hand, p2-1 is exactly divisible by 24.

I rest my case.

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• 0
First of all, any prime number squared is divisible by 24.

If you are talking about the dividion resulting in an integer, then that's true too.

By definition of prime number the square of any prime number is divisible by itself, the prime number and 1.

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• 0 p2-1 = (p+1)(p-1)

This is a remarkable little statement which is so simple and elegant that I can't believe I hadn't either seen it or discovered it before (I'm probably giving away my lack of math education, but that's ok). Thanks!

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• 0 p is prime. p>3.

1. All prime numbers are odd, p is odd

2. p^2-1=(p-1)(p+1).

3. so p^2-1 is product of two consecutive even numbers.

4. product of two consecutive even numbers are divisible by 8.

proof:

let m be odd

product=2m*(2m-2)=4(m(m-1)), as m is odd so m-1 is even and equals to 2n

so, product=8*m*n-->divisible by 8.

5. p-1,p,p+1 are three consecutive numbers. So, one of them should be divisible by 3. p is prime, so either p-1 or p+1 falls into this category.

6. Any number divisible by both 3 and 8 is divisible by 24 ## Join the conversation

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