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Thread: Prime squared minus 1

  1. #1
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    Prime squared minus 1

    I have come across this proof that for prime p >3 p^2-1 is always a multiple of 24:

    p^2 1 = (p 1)(p + 1) and given that p is prime, it must be odd.
    So p 1 and p + 1 are both even, but for two adjacent even numbers one of them must be divisible by four.

    So (p 1)(p + 1) must be divisible by eight.

    Now if p is not 3 and is prime then p can't be divisible by three, so one of (p 1) or (p + 1) must be divisible by three since every third number is divisible by three. So (p 1)(p + 1) is divisible by 3, and by 8, so it is divisible by 24.


    The two bits i bold, I am not sure about? Can anyone help me see why they are true?

    Thanks
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  2. #2
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    Re: Prime squared minus 1

    Quote Originally Posted by rodders View Post
    I have come across this proof that for prime p >3 p^2-1 is always a multiple of 24:

    p^2 – 1 = (p – 1)(p + 1) and given that p is prime, it must be odd.
    So p – 1 and p + 1 are both even, but for two adjacent even numbers one of them must be divisible by four.
    Any even number can be written as 2k for some integer k. The "next" even number is 2k+ 2= 2(k+ 1). Now, either k or k+ 1 is even. If k is even then 2k is divisible by 4. If k is not even, k+ 1 is even so 2(k+ 1) is divisible by 4.

    So (p – 1)(p + 1) must be divisible by eight.

    Now if p is not 3 and is prime then p can't be divisible by three, so one of (p – 1) or (p + 1) must be divisible by three since every third number is divisible by three.
    If p is not divisible by 3 then it is of the form p= 3k+ 1 or p= 3k+ 2. If p= 3k+ 1 then p- 1= 3k is divisible by 3. If p= 3k+ 2 then p+ 1= 3k+ 3= 3(k+ 1) so is divisible by 3

    So (p – 1)(p + 1) is divisible by 3, and by 8, so it is divisible by 24.

    The two bits i bold, I am not sure about? Can anyone help me see why they are true?

    Thanks
    Thanks from rodders
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  3. #3
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    Re: Prime squared minus 1

    That's very clear. Thank you very much!
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