# Show that $\sigma=c_1*c_1$, where $c_1$ is the constant function $1$.

I searched and wasn't able to find a question similar enough to mine. Here's the problem:

Show that $$\sigma=c_1*c_1$$, where $$c_1$$ is the constant function $$1$$.

Here is my attempt. My argument makes sense to me, but it seems kind of short. I'm wondering if there is anything I need to add to my argument or if I'm sort of using circular reasoning.

The operation $$*$$ is the convolution product defined as: $$$$\begin{split} (c_1 * c_1)(n) &= \sum_{d \mid n}c_1(d)c_1\left(\frac nd \right) \\ \end{split}$$$$

The formula above is clearly true if $$n=1$$. Assume that $$n > 1$$ and write $$n=p_1^{e_1} \dots p_s^{e_s}$$. In this sum, all terms, $$c_1(d)c_1(\frac nd)$$, are equal to $$1$$ for all $$d \mid n$$. So we will end up multiplying $$1$$ by the number of divisors, which is exactly $$\sigma(n)$$.

As always, thank you all for your help.

• You've written the question backwards, first introducing $\sigma$ and $*$, and only later telling people what you mean by those symbols. Also, $\sigma(n)$ is generally used for the sum of the divisors, not the number of divisors. – Gerry Myerson Apr 23 at 5:23