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Math Help - the number 6

  1. #1
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    the number 6

    Hi,

    Let n be a positive integer.
    Suppose that the sum of all positive divisors of n (besides n itself) is a multiple of n. (e.g. 1+2+3 = 6k where n=6 and k=1 in this case)
    Suppose too that n is a product of two distinct primes.
    Show that n must equal 6.


    I started by letting n = pq for distinct primes p and q.
    Then we know 1+p+q = kn = kpq for some k.
    This problem just seems so simple, but I can't help but think the proof is really complicated. I would appreciate any hints. Thanks.
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  2. #2
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    If p and q are distinct primes, then the sum of the positive divisors of pq other than pq is clearly 1+p+q. Hence, if 1+p+q=kpq, we would have

    p\mid q+1 and q\mid p+1

    The only pair of primes satisfying the above condition are 2 and 3.
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  3. #3
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    Quote Originally Posted by jimmyjimmyjimmy View Post
    Hi,

    Let n be a positive integer.
    Suppose that the sum of all positive divisors of n (besides n itself) is a multiple of n. (e.g. 1+2+3 = 6k where n=6 and k=1 in this case)
    Suppose too that n is a product of two distinct primes.
    Show that n must equal 6.


    I started by letting n = pq for distinct primes p and q.
    Then we know 1+p+q = kn = kpq for some k.
    This problem just seems so simple, but I can't help but think the proof is really complicated. I would appreciate any hints. Thanks.

    p \mid kpq \,\,on\,\, the \,\, RH \,\, and \,\,p \mid p \,\,\ on \,\ the \,\, LH \,\, \Longrightarrow p \mid q+1

    In the same way q \mid p+1

    So you've a pair of primes s.t. each of them divides the succesor of the other one....there aren't many of those, I believe, specially if we take into account that there's only one even prime.

    Tonio
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