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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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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\(\sigma(x)\) is the sum of positive divisors of \(x\), right?
Yes it is.
pfft too long since number theory. I really should read a book...

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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, \(\sigma(p_1^{n_1} p_2^{n_2} p_3^{n_3} \cdots p_k^{n_k})=(n_1+1)(n_2+1)(n_3+1)\cdots(n_k+1)\)
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 \(\sigma(p!)=(n_1+1)(n_2+1)(n_3+1)\cdots(n_k+1)\cdot 2 =2\sigma((p-1)!)\)
i guess this is \(\tau\) in your lexicon? and \(\sigma\) is meant to be sum of divisors instead of just counting the # of divisors?
τ(n) is the number of positive divisors of n. σ(n) is the sum of the positive divisors of n.
okay, well the proof there is for the \(\tau\) identity
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 \((1+a)(1+b)=1+a+b+ab\)
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$$\sigma(p!)=\prod_{i=0}^k\frac{p_i^{n_i+1}-1}{p_i-1}\cdot\frac{p^{1+1}-1}{p-1}=(p+1)\prod_{i=0}^k\frac{p_i^{n_i+1}-1}{p_i-1}=(p+1)\sigma((p-1)!)$$
so all these problems really require is knowledge of prime factorization and some basic algebra :p
oops, that earlier sum should be of \(p_1^{j_1}\cdots\ p_k^{j_k}\) not \(p_k^{n_k}\)
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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