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  1. #1
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    sigma function help

    I don't know where to start.

    \sigma(n) is the sum of the divisors of n, including n itself.

    Prove that there are no solutions to \sigma(n)=17.

    I was thinking of using the formula \sigma(p^a)=\frac {p^{a+1}-1}{p-1} where p is prime and a\ge1 to show that it's impossible to get 17, but I'm not sure if this is the right track.
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    Hello,
    Quote Originally Posted by Pn0yS0ld13r View Post
    I don't know where to start.

    \sigma(n) is the sum of the divisors of n, including n itself.

    Prove that there are no solutions to \sigma(n)=17.

    I was thinking of using the formula \sigma(p^a)=\frac {p^{a+1}-1}{p-1} where p is prime and a\ge1 to show that it's impossible to get 17, but I'm not sure if this is the right track.
    Let n=p_1^{a_1} \cdot p_2^{a_2} \cdots p_k^{a_k}

    Hence we know that \sigma(n)=\prod_{i=1}^k \sigma \left(p_i^{a_i}\right)=\prod_{i=1}^k \frac{p_i^{a_i+1}-1}{p_i-1}

    For \sigma(n) to be equal to 17, 17 must be equal to one of the factors, let's say \frac{p_1^{a_1+1}-1}{p_1-1}, because 17=17*1 is the only factorisation for 17.
    So it means that the others factors are equal to 1, that is to say \forall i \ne 1 \quad \frac{p_i^{a_i+1}-1}{p_i-1}=1 \quad \Rightarrow \quad a_i+1=1 \quad \text{that is to say } a_i=0. Thus n=p_1^{a_1} since all the other powers are 0. (this is an extra part lol)


    Let's go back to the problem.
    We want 17=\frac{p_1^{a_1+1}-1}{p_1-1}=1+p_1+p_1^2+\dots+p_1^{a_1}

    \implies p_1+p_1^2+\dots+p_1^{a_1}=16

    This can be factorised : p_1(1+p_1+\dots+p_1^{a_1-1})=16

    But the only prime number that divides 16 is 2. Therefore p_1=2.

    Thus 1+p_1+\dots+p_1^{a_1-1}=8, which is impossible because the left hand side is an odd number and the right hand side is an even number.


    We're done
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