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Math Help - Divisors of N

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    Divisors of N

    Show that if a 3, then an1 has at least 2v(n) divisors. (where v(n) is the number of divisors of n)
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    Quote Originally Posted by mndi1105 View Post

    Show that if a \geq 3, then a^n - 1 has at least 2\nu(n) divisors. (where \nu(n) is the number of divisors of n)
    Let d \mid n. define A(d)=\{a^d - 1, \ \frac{a^n - 1}{a^d - 1} \}. clearly A(d) is a subset of set of divisors of a^n -1. we need a trivial lemma:

    Lemma: if a \geq 3 and k \geq 1, \ \ell \geq 1, \ n \geq 1, then: (a^k - 1)(a^{\ell} - 1) \neq a^n - 1.

    Proof: suppose (a^k - 1)(a^{\ell} - 1) = a^n - 1. then we'll get: a \mid 2. \ \ \ \Box.

    so, since we've assumed that a \geq 3, the above lemma tells us that for any d \mid n, we have |A(d)|=2 and if d_1 \neq d_2, then A(d_1) \cap A(d_2) = \emptyset. \ \ \ \Box
    Last edited by NonCommAlg; October 27th 2008 at 09:32 PM.
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