p is a prime number greater than 5

Discuss the divisibility of [(p^2) - 1] by 6,12,24....

Is [(p^2) - 1] always divisible by 6, 12, 24. Yes? No? Why?

What's the logic?

Is this some kind of a series?

Is there a proof?

- Jun 27th 2010, 10:33 AMswordfish774Divisibility of [(p^2) - 1] by 6,12,24 etc. where p is prime number greater than 5
p is a prime number greater than 5

Discuss the divisibility of [(p^2) - 1] by 6,12,24....

Is [(p^2) - 1] always divisible by 6, 12, 24. Yes? No? Why?

What's the logic?

Is this some kind of a series?

Is there a proof? - Jun 27th 2010, 10:59 AMdwsmith
I would suggest a proof by induction when where k is prime for each case.

so is true

Assume is true

The question is does now - Jun 27th 2010, 11:41 AMswordfish774
Not sure, but isn't induction used when you are proving something for n = natural number. Here we have 'p' prime. So induction doesn't seem to be the appropriate method.

- Jun 27th 2010, 12:04 PMmelese
First allow me to generalize and modify your question.

Instead of a prime number , I'll consider an integer that is relatively prime to 6, and the result will imply the case for . Also, it's enought to show divisibility by 24 and then it implies divisibility by any divisor of 24, in particular 6, 12, 24.

Begin with (agree?) for some integer , then square both sides: (The signs are respectively). Finally, and either or is even, and you have divisibility by 24.

To answer your question: Yes and notice it's not special to primes geater than or equal to 5, but to all integers that are relatively prime to 6. - Jun 27th 2010, 12:17 PMmelese
- Jun 27th 2010, 12:30 PMjamix
To rearrange the question, you want to know when for

We know that for any such prime, we have and .

As it follows from the above that is always divisible by .

The really interesting question to ask is when is for n > 2.

It can be shown that if the equation has a solution then , as .

Can you figure out why?

This however doesn't identity all of the primes for which . The question on when does is pretty tough. - Jun 27th 2010, 12:38 PMchiph588@
- Jun 27th 2010, 01:04 PMjamix
Yeah, I think I missed melese's post where he/she mentioned this.

- Jun 27th 2010, 01:16 PMmelese
Isn't it impossible for to have ?

and when you have two consecutive even integers one of them is not divisible by 4, namely, has an odd divisor greater than 1.

Another thing. If , then must be a power of two itself. So a necessary (and sufficient) condition is to have for some integer , then it follows that is a Mersenne prime. Do you agree? - Jun 27th 2010, 01:30 PMswordfish774
@ melese

Brilliant! I understood. The best part was the 'prime relative to 6' part.

@jamix

I don't understand the symbols '|' 'mod' and the 3-lined symbol. Can you please elaborate.

other queries: How do you insert the equations as images? - Jun 27th 2010, 01:45 PMjamix
I meant to write and .

swordfish....

For writing math symbols, use the TEX button (see LATEX section of this forum for more details).

"mod(p)" means the remainder when an integer is divided by . For example because 2 is the remainder upon dividing 7 by 5. You could think of it as .

The sign "|" is a division sign. For example when I say what is , I'm asking for "12 divided by 3".

Does this make sense?