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Let $\varphi$ be Euler's totient function, and let $\sigma(n)=\sum_{d\mid n}d$ for $n=1,2,3,\ldots$. Both $\varphi$ and $\sigma$ are multiplicative functions. It is easy to see that the numbers $$\varphi(n^2)=n\varphi(n)\ \ (n=1,2,3,\ldots)$$ are pairwise distinct.

Question 1. Are the numbers $\varphi(n^2)\sigma(n^2)\ (n=1,2,3,\ldots)$ pairwise distinct?

Question 2. Are the rationals $\sigma(n^2)/\varphi(n^2)\ (n=1,2,3,\ldots)$ pairwise distinct?

I had the questions in 2015, and conjecture that the answers to both questions are affirmative. I have verified that $$\varphi(n^2)\sigma(n^2)\ \ (n=1,\ldots,10^7)$$ are indeed pairwise distinct, and also $$\frac{\sigma(n^2)}{\varphi(n^2)}\ \ (n=1,\ldots,10^7)$$ are pairwise distinct.

Any ideas to solve the problem? Your comments are welcome!

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    $\begingroup$ What context does this problem arise in? $\endgroup$ Commented Nov 18, 2020 at 15:33
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    $\begingroup$ In 2015 I conjectured that any positive rational number can be written as $m/n$ with $\varphi(m)$ and $\sigma(n)$ both squares. If $\varphi(m)$ and $\sigma(n)$ are both squares, then so is the product $\varphi(n)\sigma(n)$. Via computation I saw that $\varphi(n)\sigma(n)\ (n=1,2,3,\ldots)$ are not pairwise distinct. $\endgroup$ Commented Nov 19, 2020 at 1:01
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    $\begingroup$ So what is the significance of this particular conjecture? Or is it just a member of the infinite family of conjectures whether $\varphi(n^k)\sigma(n^k)$ are all pairwise distinct for $k \geq 1$? $\endgroup$ Commented Nov 19, 2020 at 7:35

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