# Understanding an inequality concerning the Euler totient function

I am trying to digest the following inequality:
$\prod_{p \mid e} \left( 1 + \frac{1}{p - 1}\right)\sum_{\substack{t \leq \frac{z}{e} \\ (t, e) = 1}} \frac{\mu^{2}(t)}{\phi(t)} \leq \sum_{d \leq z} \frac{\mu^{2}(d)}{\phi(d)}$
where $\phi$ is the Euler totient function. I am given the following hint: Note that for any squarefree $m$ dividing $e$, we have $\prod_{p \mid m} \frac{1}{p - 1} = \prod_{p \mid m} \frac{1}{\phi(p)}$. Then, using the change of variables $d = tm$, we see that if $t \leq \frac{z}{e}$ and $m \mid e$ then $tm \leq z$.
I understand we are using $\phi(p) = p - 1$ for $p$ prime. I understand that the sum of the RHS is a subset of the sum of the LHS but theres this amplification factor from the product where each product term is slightly greater than 1 and depends on the prime factorisation of $e$.
I don't understand the hint given in the reasoning specifically why must m be square free and how do we use this to deduce the main inequality.
Can someone provide a clearer explanation for this, thank you. Source: https://i.sstatic.net/cWH7H49g.png
(edited 1 month ago)
Well if m is squarefree, consider the prime factorization of m, we have
$\phi(m) = \prod_{p|m}\phi(p) = \prod_{p|m} (p-1)$,
since $\phi$ is arithmetic.
Also I'd suppose $\mu^2$ is a big hint on looking at squarefree integers anyway.

I haven't thought too deep into it though - what is $z$ anyway?
(edited 1 month ago)
Original post by tonyiptony
Well if m is squarefree, consider the prime factorization of m, we have
$\phi(m) = \prod_{p|m}\phi(p) = \prod_{p|m} (p-1)$,
since $\phi$ is arithmetic.
Also I'd suppose $\mu^2$ is a big hint on looking at squarefree integers anyway.

I haven't thought too deep into it though - what is $z$ anyway?

What you’ve written is true but I don’t see any application of it. z is just a positive integer.
the \mu^2 tells us to only consider square free t but the hint is suggesting a squarefree m implies the fact that uses \phi(p) = p - 1 but here nothing anout squarefree is required so I’m unsure where the hint wants us to make use of the squarefree assumption of m.
Original post by BrandonS15
I am trying to digest the following inequality:
$\prod_{p \mid e} \left( 1 + \frac{1}{p - 1}\right)\sum_{\substack{t \leq \frac{z}{e} \\ (t, e) = 1}} \frac{\mu^{2}(t)}{\phi(t)} \leq \sum_{d \leq z} \frac{\mu^{2}(d)}{\phi(d)}$
where $\phi$ is the Euler totient function. I am given the following hint: Note that for any squarefree $m$ dividing $e$, we have $\prod_{p \mid m} \frac{1}{p - 1} = \prod_{p \mid m} \frac{1}{\phi(p)}$. Then, using the change of variables $d = tm$, we see that if $t \leq \frac{z}{e}$ and $m \mid e$ then $tm \leq z$.
I understand we are using $\phi(p) = p - 1$ for $p$ prime. I understand that the sum of the RHS is a subset of the sum of the LHS but theres this amplification factor from the product where each product term is slightly greater than 1 and depends on the prime factorisation of $e$.
I don't understand the hint given in the reasoning specifically why must m be square free and how do we use this to deduce the main inequality.
Can someone provide a clearer explanation for this, thank you. Source: https://i.sstatic.net/cWH7H49g.png

I think you need to start by convincing yourself that

$\displaystyle \prod_{p|e} \left(1+\frac{1}{p-1} \right) = \sum_{m|e, \mu(m) \neq 0} \prod_{p|m} \frac{1}{p-1}$ (*)

noting that the sum only includes terms where m is squarefree, and making sure you understand why. After that the hint should make more sense (I don't think you actually *need* m squarefree for the later bits, but it *is* squarefree because that's how it turns out in the (*) identity).

(*) I'm pretty sure this is correct but I could be making a mistake - it's nearly 30 years since I did Number Theory.

Edit: also, your link doesn't work (site doesn't allow hotlinking. I don't know if there was more context there or anything).
(edited 1 month ago)
Original post by DFranklin
I think you need to start by convincing yourself that
$\displaystyle \prod_{p|e} \left(1+\frac{1}{p-1} \right) = \sum_{m|e, \mu(m) \neq 0} \prod_{p|m} \frac{1}{p-1}$ (*)
noting that the sum only includes terms where m is squarefree, and making sure you understand why. After that the hint should make more sense (I don't think you actually *need* m squarefree for the later bits, but it *is* squarefree because that's how it turns out in the (*) identity).
(*) I'm pretty sure this is correct but I could be making a mistake - it's nearly 30 years since I did Number Theory.
Edit: also, your link doesn't work (site doesn't allow hotlinking. I don't know if there was more context there or anything).

Re the link - if you drag to copy the web address and paste it into a new browser tab, it seems to be happy.

Can't add anything else as this isn't my area of expertise
Original post by davros
Re the link - if you drag to copy the web address and paste it into a new browser tab, it seems to be happy.
Can't add anything else as this isn't my area of expertise

Thanks - not a lot of extra context there unfortunately.

[I find the hint a bit odd; it seems to me there are 2 (arguably 3) non-trivial manipulations you need to perform to get to where you can apply the hint (and the hint seems fairly predicated on you taking that route).

Arguably the manipulations are fairly "standard manipulations when dealing with arithmetic functions" and it's just that I'm out of practice. But even then the hint seems relatively trivial].