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Math Help - Sum of powers factorization

  1. #1
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    Sum of powers factorization

    Hi everybody,

    I think that:

    <br />
N=\sum_{i=0}^{s}p^i=p^s+...+1 \quad \quad s>1<br />

    Is factorized in prime numbers (according theorem of arithmetic), that are square free.

    In other words its dinstict primes (if multiplied) give as result number N.

    i.e.
    <br />
N=p_1^{a_1}\cdot...\cdot p_k^{a_k}<br />

    where
    <br />
{a_1}=...={a_k}=1<br />

    Is this correct?

    I think this can be handled with Congruence of Clausen and von Staudt proof. Unfortunatelly i cannot find any link (with proof) into internet.

    Thank you a lot
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    You mean that 1+p+...+p^s=\frac{p^{s+1}-1}{p-1} is always squarefree?

    Or does p have to be prime?
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    In any case a counter-example is 1+7+7^2+7^3=2^45^2
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  4. #4
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    Ok counter example is definite thanks
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    I suggest always checking a couple of cases before conjecturing something, let alone trying to prove it. Counter-examples to your conjecture abound. It took me about three tries to find one.
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