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Math Help - congruence

  1. #1
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    congruence

    find all x such that x^3-x+1 is congruent to 0 mod 6.

    i tried but it seems like there is no x
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  2. #2
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    Re: congruence

    You are right.

    0^3- 0+ 1= 1
    1^3- 1+ 1= 1
    2^3- 2+ 1= 7= 1 mod 6
    3^3- 3+ 1= 25= 1 mod 6
    4^3- 4+ 1= 61= 1 mod 6
    5^3- 5+ 1= 121= 1 mod 6
    6^3- 6+ 1= 211= 1 mod 6
    That's all you have to do. For all x, x^3- x+ 1= 1 mod 6.
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  3. #3
    Member Sylvia104's Avatar
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    Re: Congruence

    A little observation should tell you that x^3-x+1 is always odd and therefore can never be divisible by 6.

    If you delve further, you have x^3-x=(x-1)x(x+1), a product of three consecutive integers this is always divisible by 6 because it always contains exactly one multiple of 3 and at least one multiple of 2. Hence x^3-x+1\equiv1\mod6 for all x\in\mathbb Z.
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