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Homework Help: How to make the Tau function odd

  1. May 11, 2007 #1
    1. The problem statement, all variables and given/known data
    What are the necessary conditions on n for T(n) {the Tau function} to be odd.

    2. Relevant equations
    Tau function counts the number of divisors of n
    T(1) = 1, T(p) = 2, and T(p^k) = k + 1

    3. The attempt at a solution
    I have that n cannot be prime, also if n = p^k then k must be even or of the from k = 2c for some c > 0.

    I cannot figure out how to describe evens with were are multiple primes such as 1x3x7x8 = 231. That number has an even number of divisors but I'm and lost at how to prevent that.
    Last edited: May 11, 2007
  2. jcsd
  3. May 11, 2007 #2

    matt grime

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    I don't understand why k is even OR k=2c. Those are identical conditions....

    What is T(xy) for x,y coprime?
  4. May 11, 2007 #3
    I meant k is even i.e k =2c... I was just adding more definition.

    T(xy) for x,y is coprime is an even number of divisors. Thanks.
  5. May 11, 2007 #4


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    No, that's not right.
  6. May 11, 2007 #5
    T(xy) for x,y is coprime means T(x)T(y)
  7. May 11, 2007 #6
    Is is sufficient to say that T(n) is only odd when n is a perfect square?
  8. May 11, 2007 #7


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    T(n) is odd if and only if n is a perfect square. Can you prove it?
  9. May 11, 2007 #8
    I was just supposed to make a conjecture, but I'll try to work on a proof
  10. May 11, 2007 #9
    Assume that n is an odd tau number. Let n = p^a1*p^a2...* p^ak. By the definition of tau number (a1 + 1)(a2 + 1)(a3 + 1)..... (ak + 1) | n. Therefore for any 0 < i < k + 1, ai + 1 is odd, and hence ai is even. Since every prime in the factorization of n is raised to an even power, n is a perfect square.
  11. May 12, 2007 #10


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    What you meant to say was

    [tex](a_1 + 1)(a_2 + 1)(a_3 + 1)\dots (a_k + 1) = \tau (n)[/tex]


    [tex](a_1 + 1)(a_2 + 1)(a_3 + 1)\dots (a_k + 1) | n[/tex]
    The question only asked you to find a necessary condition, and you've done this. But the stuff you've said above goes both ways, so it's actually a sufficient condition as well.
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