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

Question

$\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$, where $\tau(d)$ designates the number of positive divisors of d.

Now I only know that both sides are multiplicative functions, could you tell me what I need to do next?

Answer 1

If you've shown that both the LHS and the RHS are multiplicative functions, then you must now show it's true for arbitrary prime powers $n=p^r$. In doing so, use $\tau(n)=\sigma_1(p^r)=r+1$ and this. It then follows for all composite numbers by prime factorizing both sides through the multiplication.

Answer 2

Recall that if $f$ is a multiplicative function, and $$g(n)=\sum_{d|n}f\left(\frac{d}{n}\right),$$ then $g$ is a multiplicative function.

From this you can deduce that both sides are multiplicative functions. Thus to verify the identity, all you need to do is to verify it for $n=p^k$, where $p$ is prime.

For $n=p^k$, the left-hand side is $1^3+2^3+\cdots +k^3$. The right-hand side is $(1+2+\cdots+k)^2$. The fact that the two are equal is a probably familiar fact. If it isn't, it can be proved by induction.

Answer 3

To get this result you need to show

(a) it is true for prime powers, which you can show with $$ \sum_1^n i^3 = \frac{n^2(n+1)^2}{4} = \left(\sum_1^n i\right)^2$$

(b) both the left and right sides are multiplicative, which you say you know