- #1

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

Thanks!

- Thread starter AxiomOfChoice
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- #1

- 533

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

Thanks!

- #2

Hurkyl

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Compute the greatest integer that divides [itex]p^4 - 1[/itex] for every prime number *p* in the range 5 < p < N

where

- #3

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Wouldn't it be [tex]p^4-1[/tex]? Maybe I'm not understanding the question.What is the greatest integer that divides [itex]p^4 - 1[/itex] for every prime number [itex]p[/itex] greater than 5?"

- #4

- 533

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(A) 12

(B) 30

(C) 48

(D) 120

(E) 240

- #5

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[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.

- #6

HallsofIvy

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Yes, you are. The question is about aWouldn't it be [tex]p^4-1[/tex]? Maybe I'm not understanding the question.

- #7

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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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