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Math Help - Mod proof with primes

  1. #1
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    Mod proof with primes

    If p is greater than or equal to 5 and p is prime, prove that [p]=[1] or [p]=[5] in Z6.

    I think that we will have to consider cases, that is,

    p=6q, p=6q+1, p=6q+2, p=6q+3, p=6q+4, p=6q+5....

    Since p is prime, we have p divides p and 1 divides p....Also, since p is greater than or equal to 5, it must be odd, i.e. like 2k+1, but I don't know if this is useful.

    I really don't know what I need to show this....
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  2. #2
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    Quote Originally Posted by twittytwitter View Post
    If p is greater than or equal to 5 and p is prime, prove that [p]=[1] or [p]=[5] in Z6.

    I think that we will have to consider cases, that is,

    p=6q, p=6q+1, p=6q+2, p=6q+3, p=6q+4, p=6q+5....

    Since p is prime, we have p divides p and 1 divides p....Also, since p is greater than or equal to 5, it must be odd, i.e. like 2k+1, but I don't know if this is useful.

    I really don't know what I need to show this....

    If [p]=0,2,4\!\!\!\pmod 6 then p is even, if...etc.

    Tonio
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  3. #3
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    Quote Originally Posted by twittytwitter View Post
    If p is greater than or equal to 5 and p is prime, prove that [p]=[1] or [p]=[5] in Z6.

    I think that we will have to consider cases, that is,

    p=6q, p=6q+1, p=6q+2, p=6q+3, p=6q+4, p=6q+5....

    Since p is prime, we have p divides p and 1 divides p....Also, since p is greater than or equal to 5, it must be odd, i.e. like 2k+1, but I don't know if this is useful.

    I really don't know what I need to show this....

    Hey, did you prove this one....
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