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

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

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If you have proved that both sides are multiplicative functions, it suffices to prove that the statement holds for powers of primes. In this case, it is straight forward to compute the value of each side. It may be useful to use the fact that $\sum_{i=1}^n i^3=(\sum_{i=1}^n i)^2$. – Aaron Nov 3 '11 at 0:06

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.

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

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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

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