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
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?
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.
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.
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?
I meant to write and .
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?