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Thread: 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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  2. #2
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    Quote Originally Posted by mndi1105 View Post

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

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

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

    so, since we've assumed that $\displaystyle a \geq 3,$ the above lemma tells us that for any $\displaystyle d \mid n,$ we have $\displaystyle |A(d)|=2$ and if $\displaystyle d_1 \neq d_2,$ then $\displaystyle A(d_1) \cap A(d_2) = \emptyset. \ \ \ \Box$
    Last edited by NonCommAlg; Oct 27th 2008 at 08:32 PM.
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