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anonymous

  • one year ago

For any prime p, establish each of the following assertions below: \[\sigma (p! )=(p+1)\sigma((p-1)!)\] \[\tau (p! )=2\tau((p-1)!)\]

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  1. anonymous
    • one year ago
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    \(\sigma(x)\) is the sum of positive divisors of \(x\), right?

  2. anonymous
    • one year ago
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    Yes it is.

  3. zzr0ck3r
    • one year ago
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    pfft too long since number theory. I really should read a book...

  4. anonymous
    • one year ago
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    .

  5. anonymous
    • one year ago
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    well, it's trivial to show that the divisors of \(n=p_1^{n_1} p_2^{n_2} p_3^{n_3}\cdots \ p_k^{n_k}\) must be of the form \(p_1^{j_1} p_2^{j_2} p_3^{j_3} \cdots \ p_k^{j_k}\) where \(0\le j_i\le n_i\); in other words, there are \(n_i+1\) choices for the exponent of \(p_i\) and so a total \((n_1+1)(n_2+1)(n_3+1)\cdots(n_k+1)\) such combinations; in other words, \(\sigma(p_1^{n_1} p_2^{n_2} p_3^{n_3} \cdots p_k^{n_k})=(n_1+1)(n_2+1)(n_3+1)\cdots(n_k+1)\)

  6. anonymous
    • one year ago
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    now, if \(p\) is prime then \(p\not|\ (p-1)!\); in other words, \((p-1)!\) as a prime factorization can be written \((p-1)!=p_1^{n_1} p_2^{n_2} p_3^{n_3}\cdots\ p_k^{n_k}\) where \(p_i\ne p\), so then \(p!=p(p-1)!=p_1^{n_1} p_2^{n_2} p_3^{n_3}\cdots\ p_k^{n_k} p^1\) so it follows \(\sigma(p!)=(n_1+1)(n_2+1)(n_3+1)\cdots(n_k+1)\cdot 2 =2\sigma((p-1)!)\)

  7. anonymous
    • one year ago
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    i guess this is \(\tau\) in your lexicon? and \(\sigma\) is meant to be sum of divisors instead of just counting the # of divisors?

  8. anonymous
    • one year ago
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    τ(n) is the number of positive divisors of n. σ(n) is the sum of the positive divisors of n.

  9. anonymous
    • one year ago
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    okay, well the proof there is for the \(\tau\) identity

  10. anonymous
    • one year ago
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    for the sum of the positive divisors of \(p_1^{n_1} p_2^{n_2} p_3^{n_3}\cdots\ p_k^{n_k}\) it's clearly $$\sum_{j_1=0}^{n_1}\sum_{j_2=0}^{n_2}\sum_{j_3=0}^{n_3}\cdots\sum_{j_k=0}^{n_k} p_1^{j_1}p_2^{j_2}p_3^{j_3}\cdots p_k^{n_k}=\prod_{i=0}^k\left(\sum_{j=0}^{n_i} p_i^j\right)=\prod_{i=0}^k\frac{p_i^{n_i+1}-1}{p_i-1}$$ using the fact \((1+a)(1+b)=1+a+b+ab\)

  11. anonymous
    • one year ago
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    then the argument is pretty much identical form: $$(p-1)!=p_1^{n_1}\cdots p_k^{n_k}\\p!=p(p-1)!=p_1^{n_1}\cdots p_k^{n_k} p^1$$so$$\sigma(p!)=\prod_{i=0}^k\frac{p_i^{n_i+1}-1}{p_i-1}\cdot\frac{p^{1+1}-1}{p-1}=(p+1)\prod_{i=0}^k\frac{p_i^{n_i+1}-1}{p_i-1}=(p+1)\sigma((p-1)!)$$

  12. anonymous
    • one year ago
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    so all these problems really require is knowledge of prime factorization and some basic algebra :p

  13. anonymous
    • one year ago
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    oops, that earlier sum should be of \(p_1^{j_1}\cdots\ p_k^{j_k}\) not \(p_k^{n_k}\)

  14. anonymous
    • one year ago
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    note that this identity gives us \(\sigma(p!)=(p+1)\sigma((p-1)!)=\dots=(p+1)!\cdot \sigma(1)=(p+1)!\)

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