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Math Help - product of primes II

  1. #1
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    product of primes II

    Let q_i, p be odd primes, k \ge 2 integer

    If \prod_{i=1}^{k}q_i = p^5+2
    then

     <br />
\prod_{i=1}^{k}(q_i-1) \not= p^5-p^4+2<br />
    Last edited by wauwau; December 27th 2009 at 07:17 AM.
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  2. #2
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    <br />
\prod_{i=1}^{k}q_i = q_1 q_2 \cdot\cdot\cdot q_k = p^5+2<br />

    <br />
\prod_{i=1}^{k}(q_i-1) = (q_1-1) (q_2-1) \cdot\cdot\cdot (q_k-1) = p^5 - p^4 +2<br />


     \phi(q_1 q_2 \cdot\cdot\cdot q_k) = (q_1 - 1) (q_2 - 1) \cdot \cdot \cdot (q_k - 1) (I'll leave this proof to someone else)

    So, let's consider the following, seeking contradiction

     \phi(p^5+2) = p^5 - p^4 + 2 .


    Quite honestly, I'm not sure if this is how you attack this. Fermat's little Thm. applied to  P^5 sheds light on the fact that
     p^5 \equiv p (\mod 5) \implies p^5 - p \equiv 0 (mod 5) \implies (p-1)(p)(p+1)(p^2+1) \equiv 0 (mod 5)
    .

    Here we observe that this  p^5 - p = (p-1)(p)(p+1)(p^2+1) bit is divisible by 2 (2 consecutive numbers multiplied together), divisible by 3 (one of three consecutive terms certainly is divisible by 3), divisible by 5 (by virtue of the 0(mod 5)), and divisible by 6 since that's 2*3.

    Just offering my $0.02. I'd be interested in a elegant solution to this.

    -Andy
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