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Question about primes and divisibility abstract algebra/number theory

  1. Apr 12, 2009 #1
    Can someone please tell me how to go about answering a question like this? I've been racking my brain for a long time and still don't have a clue...I guess because my background in algebra/number theory really isn't that strong.

    "What is the greatest integer that divides [itex]p^4 - 1[/itex] for every prime number [itex]p[/itex] greater than 5?"

    Thanks!
     
  2. jcsd
  3. Apr 12, 2009 #2

    Hurkyl

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    There seems an obvious first thing to try:
    Compute the greatest integer that divides [itex]p^4 - 1[/itex] for every prime number p in the range 5 < p < N​

    where N is whatever number you like. I'd probably start with 10 and then increase it a few times until I had an idea what was going on.
     
  4. Apr 12, 2009 #3
    Wouldn't it be [tex]p^4-1[/tex]? Maybe I'm not understanding the question.
     
  5. Apr 12, 2009 #4
    Sorry; this is a multiple choice question off of an old Math Subject GRE exam. There are five answer choices:

    (A) 12
    (B) 30
    (C) 48
    (D) 120
    (E) 240
     
  6. Apr 12, 2009 #5
    This is what I have so far.

    [tex]p^4-1= (p+1)(p-1)(p^2+1)[/tex]

    p is odd so [tex]p = 1 \text{ or } 3 [/tex] (mod 4) so there are three 2's in (p+1) and (p-1) plus another in [tex](p^2+1)[/tex] so [tex]16|p^4-1[/tex]. Furthermore, 3 does not divide p (since p>5) so (p-1) or (p+1) does and so [tex]3|p^4-1[/tex]. Now it's between 240 and 48.
     
  7. Apr 12, 2009 #6

    HallsofIvy

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    Yes, you are. The question is about a single number that divides [itex]p^4- 1[/itex] for all primes p> 5. It cannot depend on p.
     
  8. Apr 12, 2009 #7
    Alright I found the the last factor.

    1^2 = 1 mod 5
    2^2 = 4
    3^2 = 4
    4^2 = 1


    So [tex]p^2[/tex] = 1 or 4 mod 5

    [tex](p^2)^2 = 1[/tex] mod 5

    [tex]p^4-1 = 0[/tex] mod 5
     
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