Let's have some fun with Mphlee's introduction of radicals. The radical we care about is:
\[
\sqrt{\mathcal{I}_m} = \{n \in \mathbb{N}\,\mid\, m = n^k\,\,k \in \mathbb{N}\}
\]
We are then writing:
\[
\sum_{n \le m} \chi_m(n) = \sum_{n \in \sqrt{\mathcal{I}_m}} 1\\
\]
This becomes the quicker statement of:
\[
\sum_{n \le m} \chi_m(n) = \left| \sqrt{\mathcal{I}_m}\right|\\
\]
Now this is all super cool and all. But it's just a rewrite--a translation from the language I was using. But here, I can better explain my solution:
\[
|\sqrt{\mathcal{I}_m}| \le \Pi(m)\\
\]
Where \(\Pi(m)\) is a much more manageable quantity... If:
\[
\begin{align}
m &= p_1^{r_1} p_2^{r_2} \cdots p_c^{r_c}\\
\Pi(m) &= \sigma(r_1) \sigma (r_2) \cdots \sigma(r_c)\\
\end{align}
\]
This does not prove the size of \(|\sqrt{\mathcal{I}_m}|\), or even really discuss the radical. All we know, is that there's a pretty good bound of \(\Pi(m)\)...
I'd love to do this radical talk, Mphlee. But I don't see anything obvious that would aid in the questions at hand. It's just a different way of saying the problem (which is still cool), but nothing novel to the problem.
Unless this radical produces a brand new multplication formula or something, I don't give a fuck......
Nothing but love, Vittorio. Just talking straight
\[
\sqrt{\mathcal{I}_m} = \{n \in \mathbb{N}\,\mid\, m = n^k\,\,k \in \mathbb{N}\}
\]
We are then writing:
\[
\sum_{n \le m} \chi_m(n) = \sum_{n \in \sqrt{\mathcal{I}_m}} 1\\
\]
This becomes the quicker statement of:
\[
\sum_{n \le m} \chi_m(n) = \left| \sqrt{\mathcal{I}_m}\right|\\
\]
Now this is all super cool and all. But it's just a rewrite--a translation from the language I was using. But here, I can better explain my solution:
\[
|\sqrt{\mathcal{I}_m}| \le \Pi(m)\\
\]
Where \(\Pi(m)\) is a much more manageable quantity... If:
\[
\begin{align}
m &= p_1^{r_1} p_2^{r_2} \cdots p_c^{r_c}\\
\Pi(m) &= \sigma(r_1) \sigma (r_2) \cdots \sigma(r_c)\\
\end{align}
\]
This does not prove the size of \(|\sqrt{\mathcal{I}_m}|\), or even really discuss the radical. All we know, is that there's a pretty good bound of \(\Pi(m)\)...
I'd love to do this radical talk, Mphlee. But I don't see anything obvious that would aid in the questions at hand. It's just a different way of saying the problem (which is still cool), but nothing novel to the problem.
Unless this radical produces a brand new multplication formula or something, I don't give a fuck......
Nothing but love, Vittorio. Just talking straight
