# Help proving an equality about arithmetic functions

So I have this equality to prove for any $n \in \mathbb{N}$:

$$\sum_{d|n} \sigma(d) \phi(\frac{n}{d}) = n \tau(n)$$

So I was able to show that left and right side are multiplicative.

So how can I prove that the left side is equal to the right.

So if it holds for $n=1$, $n=2$

Then I take:

$n=p_1^{\alpha_1}...p_k^{\alpha_k}$ And have to see if it works for this.

But am not sure how do I represent the left side, since I don't know how this actually affects the sum $\phi(\frac{n}{d})$.

Any help, answers or hints would be appreciated.

• @DanielFischer So only set $n=p^k$ and then check it ? we only solved in class such examples where we used always this: $n=p_1^{\alpha_1}...p_k^{\alpha_k}$ is it the same, or what is the principle behind using only $n=p^k$. I apologize for the long reply, but just making sure i can use this. – MathIsTheWayOfLife Jun 21 '18 at 17:36
• If $f$ is a multiplicative function and $n = p_1^{\alpha_1}\cdot \dotsc \cdot p_k^{\alpha_k}$, then $$f(n) = \prod_{\kappa = 1}^k f(p_{\kappa}^{\alpha_{\kappa}})\,.$$ Thus if $f$ and $g$ are two multiplicative functions, then $f(p^{\alpha}) = g(p^{\alpha})$ for all primes $p$ and $\alpha \in \mathbb{N}$ already implies $f(n) = g(n)$ for all $n$. Typically, it's much easier to check the equality at prime powers than at general $n$, so that saves some work. – Daniel Fischer Jun 21 '18 at 17:41