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Greatest integer divides p^4 -1

  1. Dec 20, 2003 #1
    What is the greatest integer divides p^4 -1 for every prime number p greater than 5?

    It is 240. Why?

    Thanks!!
     
  2. jcsd
  3. Dec 20, 2003 #2

    Hurkyl

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    Try factoring [itex]p^4-1[/itex].
     
  4. Dec 20, 2003 #3
    I already figured this out. (p^4-1) = (p^2+1)(p-1)(p+1).
    i.) Each term is divisible by 2 since p is odd. Also, either p-1 or p+1 is divisible by 4.
    So divisible by 16.
    ii.) either p mod 3 = 1 or p mod 3 = 2. If the first case, p-1 = 0 mod 3, second case, p^2 -1 = 0 mod 3.
    So divisible by 3.
    iii.) either p mod 5 = 1, p mod 5 = 2, p mod 5 = 3, or p mod 5 = 4.
    If p mod 5 = 1, p-1 = 0 mod 5. If p mod 4 = 1, p+1 = 0 mod 5, and if p mod 5 = 2 or p mod 3 = 1, then p^2+1 = 0 mod 5.
    So divisible by 5.

    Hence, divisible by 16*3*5 = 240.
     
  5. Dec 20, 2003 #4

    Hurkyl

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    Er, my mistake, I misread the problem.

    Well, you've verified that [itex]240 | p^4 - 1[/itex] for any prime number greater than 5, correct? (In fact, [itex]240 | n^4 - 1[/itex] if [itex](240, n) = 1[/itex])

    The easiest way to proceed from here is, I think, to start looking at some explicit examples, and finish the proof from a small number of those. For instance, if [itex]m | p^4 - 1[/itex] for all primes [itex]p > 5[/itex], then [itex]m | 7^4 - 1 = 2400[/itex].
     
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