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How do I see this? Simplification

  1. Jul 27, 2011 #1
    How do I see that for any natural non-zero n:

    [itex]\left( \prod_{1 \leq d \leq n \quad d | n} \right) \left( \prod_{0 \leq i \leq d-1 \quad \textrm{gcd}(d,i)=1} \right) a_i = \prod_{0 \leq i \leq n-1} a_i[/itex]?
  2. jcsd
  3. Jul 28, 2011 #2
    I would say this is going to be hard to prove. EXCEEDINGLY hard to prove. Because... it's not true. For instance, letting n = 6, the LHS is

    a_0 a_1^3 a_2 a_5

    The RHS, of course, is

    a_0 a_1 a_2 a_3 a_4 a_5

    Or maybe I don't understand your notation?
  4. Jul 28, 2011 #3
    Well that explains why I couldn't prove it :p

    I was (am) trying to prove that [itex]X^n - 1 = \left( \prod_{1 \leq d \leq n \quad d | n} \right) \Phi_d[/itex] where [itex]\Phi_d = \left( \prod_{0 \leq i \leq d-1 \quad \textrm{gcd}(d,i)=1} \right) (X - \omega_d^i)[/itex] where [itex]\omega_d = \exp{ (2 \pi i /d) }[/itex]

    The RHS of my original equation can be understood by noting that [itex]X^n-1=\prod_{0 \leq i \leq n} (X-\omega_n^i)[/itex]

    but I now see my error... My apologies

    However, I'm still not sure how to prove the above equation, aka that

    [itex]\prod_{0 \leq i \leq n} (X-\omega_n^i) = \left( \prod_{1 \leq d \leq n \quad d | n} \right) \left( \prod_{0 \leq i \leq d-1 \quad \textrm{gcd}(d,i)=1} \right) (X - \omega_d^i)[/itex]
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