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anonymous
 one year ago
For any prime p, establish each of the following assertions below:
\[\sigma (p! )=(p+1)\sigma((p1)!)\]
\[\tau (p! )=2\tau((p1)!)\]
anonymous
 one year ago
For any prime p, establish each of the following assertions below: \[\sigma (p! )=(p+1)\sigma((p1)!)\] \[\tau (p! )=2\tau((p1)!)\]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\sigma(x)\) is the sum of positive divisors of \(x\), right?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0pfft too long since number theory. I really should read a book...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, 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, \(\sigma(p_1^{n_1} p_2^{n_2} p_3^{n_3} \cdots p_k^{n_k})=(n_1+1)(n_2+1)(n_3+1)\cdots(n_k+1)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now, if \(p\) is prime then \(p\not\ (p1)!\); in other words, \((p1)!\) as a prime factorization can be written \((p1)!=p_1^{n_1} p_2^{n_2} p_3^{n_3}\cdots\ p_k^{n_k}\) where \(p_i\ne p\), so then \(p!=p(p1)!=p_1^{n_1} p_2^{n_2} p_3^{n_3}\cdots\ p_k^{n_k} p^1\) so it follows \(\sigma(p!)=(n_1+1)(n_2+1)(n_3+1)\cdots(n_k+1)\cdot 2 =2\sigma((p1)!)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i guess this is \(\tau\) in your lexicon? and \(\sigma\) is meant to be sum of divisors instead of just counting the # of divisors?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0τ(n) is the number of positive divisors of n. σ(n) is the sum of the positive divisors of n.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, well the proof there is for the \(\tau\) identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 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 $$\sum_{j_1=0}^{n_1}\sum_{j_2=0}^{n_2}\sum_{j_3=0}^{n_3}\cdots\sum_{j_k=0}^{n_k} p_1^{j_1}p_2^{j_2}p_3^{j_3}\cdots p_k^{n_k}=\prod_{i=0}^k\left(\sum_{j=0}^{n_i} p_i^j\right)=\prod_{i=0}^k\frac{p_i^{n_i+1}1}{p_i1}$$ using the fact \((1+a)(1+b)=1+a+b+ab\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then the argument is pretty much identical form: $$(p1)!=p_1^{n_1}\cdots p_k^{n_k}\\p!=p(p1)!=p_1^{n_1}\cdots p_k^{n_k} p^1$$so$$\sigma(p!)=\prod_{i=0}^k\frac{p_i^{n_i+1}1}{p_i1}\cdot\frac{p^{1+1}1}{p1}=(p+1)\prod_{i=0}^k\frac{p_i^{n_i+1}1}{p_i1}=(p+1)\sigma((p1)!)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so all these problems really require is knowledge of prime factorization and some basic algebra :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, that earlier sum should be of \(p_1^{j_1}\cdots\ p_k^{j_k}\) not \(p_k^{n_k}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0note that this identity gives us \(\sigma(p!)=(p+1)\sigma((p1)!)=\dots=(p+1)!\cdot \sigma(1)=(p+1)!\)
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