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Thread: Prime squared -1 problem

  1. #1
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    Prime squared -1 problem

    If I square a prime number ( >3) and subtract 1 , it turns out it is a multiple of 24 and there are different ways of proving this.
    Why doesn't my approach below work?

    All prime number are of the form 6n+1 or 6n-1 ( greater than 3).
    So (6n+1)^2 -1 or (6n-1)^2 -1 only proves that they are multiples of 12 not 24?

    What am I missing?
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  2. #2
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    Re: Prime squared -1 problem

    $(6n+1)^2 -1 = 36n^2+12n = 12(3n^2+n)$

    if n is odd ...

    $12[3(2k+1)^2 + (2k+1)]=12(12k^2+14k+4)=24(6k^2+7k+2)$

    if n is even ...

    $12[3(2k)^2+(2k)]=12(12k^2+2k)=24(6k^2+k)$


    you can show the same for the other prime form ...
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  3. #3
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    Re: Prime squared -1 problem

    (6n+1)^2- 1= 36n^2+ 12n= 12n(3n+ 1).

    Obviously that has a factor of 12 but what about n(3n+ 1)? If n is even then there is another factor of 2 so 2*12= 24 is a factor. If n is odd, then 3n is also odd so 3n+ 1 is even and, again, there is another factor of 2. 24 is always a factor.
    Last edited by HallsofIvy; Jun 16th 2017 at 05:18 AM.
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  4. #4
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    Re: Prime squared -1 problem

    Of course. Thank you both!
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