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Thread: Nonnegative integers problem

  1. #1
    Junior Member
    Oct 2009

    Nonnegative integers problem

    Let N=30030, which is the product of the first six primes. How many nonnegative integers x less than N have the property that N divides x^3-1?
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  2. #2
    Aug 2009
    I'm not sure how to do this in a more direct way ... here's what I got so far:


    So, N|x-1 or N|x^2+x+1

    The first case gives us x=1, and nothing else since x<30030

    The second case has no solutions, and I may be wrong here somewhere (it's late in my part of the world )

    N|x(x+1) + 1, since N is the product of the first 6 primes, this includes 2 ... so 2|x(x+1) + 1. But, x(x+1) will be even, and by adding 1, you get an odd number, so 2 can't divide x(x+1) + 1.

    So ... I believe the answer is that there is only 1 solution, which is when x=1
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