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Thread: Number Theory - Primes and Eulers equation

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    Number Theory - Primes and Eulers equation

    I really hate these types of questions. Can anyone help please?
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    1. Given $\displaystyle n>1$ in prime factorized form $\displaystyle n=p_1^{a_1}...p_m^{a_m}$ then the number of divisors is $\displaystyle (a_1+1)...(a_m+1)$.

    Use this to try to find the answer.
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    2. Since $\displaystyle n$ is prime it means $\displaystyle \sigma (n) = n+1$. Thus, $\displaystyle n+2\sigma (n) = n+2(n+1)$. Show this is not divisible by three by contradiction.


    If $\displaystyle n=p^2$ then $\displaystyle \sigma (p^2)=1+p+p^2$. And thus, $\displaystyle n+2\sigma (n) = 3p^2+2p+2 \equiv 2p+2 \equiv p+1\equiv 0 (\bmod 3)$.

    Try to do the last one by substituting $\displaystyle \sigma (pq) = 1+p+q+pq$ into the expression and seeing what you get.
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    3. We can write $\displaystyle n=p^a m$ where $\displaystyle p\not | m \mbox{ and }a\geq 2$. Thus, $\displaystyle \gcd (p_a,m)=1$.

    Then,
    $\displaystyle \phi (n) = \phi (p^a) \phi (m)$
    $\displaystyle \phi \left( \frac{n}{p} \right) = \phi (p^{a-1})\phi(m)$

    Use the above equation to reach the relationship you are trying to show.
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    Quote Originally Posted by ThePerfectHacker View Post
    2. Since $\displaystyle n$ is prime it means $\displaystyle \sigma (n) = n+1$. Thus, $\displaystyle n+2\sigma (n) = n+2(n+1)$. Show this is not divisible by three by contradiction.


    If $\displaystyle n=p^2$ then $\displaystyle \sigma (p^2)=1+p+p^2$. And thus, $\displaystyle n+2\sigma (n) = 3p^2+2p+2 \equiv 2p+2 \equiv p+1\equiv 0 (\bmod 3)$.

    Try to do the last one by substituting $\displaystyle \sigma (pq) = 1+p+q+pq$ into the expression and seeing what you get.
    I'm curious. What is the $\displaystyle \sigma$ function?

    -Dan
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    Quote Originally Posted by topsquark View Post
    I'm curious. What is the $\displaystyle \sigma$ function?

    -Dan
    The SUM of all divisors.
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