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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

$$\sigma(x)$$ is the sum of positive divisors of $$x$$, right?

2. anonymous

Yes it is.

3. zzr0ck3r

pfft too long since number theory. I really should read a book...

4. anonymous

.

5. anonymous

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

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

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

τ(n) is the number of positive divisors of n. σ(n) is the sum of the positive divisors of n.

9. anonymous

okay, well the proof there is for the $$\tau$$ identity

10. anonymous

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

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

so all these problems really require is knowledge of prime factorization and some basic algebra :p

13. anonymous

oops, that earlier sum should be of $$p_1^{j_1}\cdots\ p_k^{j_k}$$ not $$p_k^{n_k}$$

14. anonymous

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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