# A couple proofs.

• May 1st 2008, 08:09 AM
Proof_of_life
A couple proofs.
Prove that:

(1) $\sum_{d|n}\tau^3(d) = \left(\sum_{d|n}\tau(d)\right)^2$

(2) $\prod_{d|n}d = n^{\frac{\tau(n)}{2}}$

These come from my Arithmetic functions section in my book. $\tau$ is a divisor function and is multiplicative.

Thanks for any help!
• May 1st 2008, 09:09 AM
ThePerfectHacker
Quote:

Originally Posted by Proof_of_life
(1) $\sum_{d|n}\tau^3(d) = \left(\sum_{d|n}\tau(d)\right)^2$

Let $f(n) = \sum_{d|n} \tau^3(d)$ and $g(n) = \left( \sum_{d|n}\tau (d) \right)^3$.

The functions $f(n),g(n)$ are weakly multiplicative meaning $f(ab)=f(a)f(b) \text{ and }g(ab) = g(a)g(b)$ whenever $\gcd(a,b) = 1$.

This means it remains to prove $f(p^k) = g(p^k)$ for all primes $p$ and positive integers $k$.

$f(p^k) = \tau^3 (1) + \tau^3 (p) + ... + \tau^3 (p^k) = 1^3 + 2^3 + ... + (k+1)^3$
$g(p^k) = ( 1 + \tau (p) + ... + \tau (p^k) )^2 = (1+2+...+(k+1))^2$.

This if of course true using the identity, $(1+2+...+n)^2 = 1^3+2^2+...+n^3$.

Quote:

(2) $\prod_{d|n}d = n^{\frac{\tau(n)}{2}}$
If $n$ is not a square then among its factors $d_1,d_2,...,d_m$ for any $d_i$ we can find a mate $d_j$ so that $d_id_j=n$. Thus, we get $n^{\tau(n)/2}$. If $n$ is a square we need to modify the proof a little bit.

This is Mine 95:):)th Post!!!
• May 1st 2008, 09:31 AM
PaulRS
We have: $\sum_{d|n}{\ln(d)}=\sum_{d|n}{\ln\left(\frac{n}{d} \right)}$

$\sum_{d|n}{\ln(d)}=\sum_{d|n}{\ln(n)}-\sum_{d|n}{\ln(d)}$

Thus: $2\cdot{\sum_{d|n}{\ln(d)}}=\tau(n)\cdot{\ln(n)}$

And then (2) follows easily