In this post we study the following equations that involve quasiperfect numbers, denoted as $x$, that are integers such that the sum of all its positive divisors is equals to $2x+1$, and certain multiplicative functions These are the sum of divisors function $\sum_{1\leq d \mid n}d$ denoted as $\sigma(n)$, the Euler's totient function $\varphi(n)$ and the product of distinct prime numbers dividing an integer $n>1$ that is denoted $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p\,.$$ The spirit of the post was to create equations involving these arithmetic functions, and now I ask if you can to find odd solutions of these equations (since I think that ask about what work can be done about these equations is very difficult to solve, I prefer in this ocassion ask about possible odd solutions, thus I present the following very similar equations/questions in which you can to find an odd solution you as a counterexample or well discuss from a theoretical/heuristic point of view if one can to hope odd solutions of our equations).

Thus a characterization of quasiperfect numbers is given as $$\sigma(x)=2x+1,$$

you can see for example the Wikipedia with title Quasiperfect number as a reference for this post.

Fact. If $x$ is a quasiperfect number then the following identities are satisfied $$\varphi(x^{\sigma(x)}\sigma(x)^x)=x^{2x}(2x+1)^{x-1}\varphi(x)\varphi(2x+1),\tag{1}$$ $$\sigma(x^{\sigma(x)}\sigma(x)^x)=\sigma\left(x^{2x+1}\right)\sigma((2x+1)^x)\tag{2}$$ and $$\operatorname{rad}\left(\frac{\varphi(x^{\sigma(x)}\sigma(x)^x)}{\varphi(x(2x+1))}\right)=\operatorname{rad}(x)\operatorname{rad}(2x+1).\tag{3}$$

As aside remark, one can also multiply by $f(2^{\alpha})$ previous equations, where $\alpha$ is a positive integer and $f(n)$ the corresponding multiplicative function in each identity.

Question. Can you find an odd integer $1\leq X$ satisfying the equation $(1)$, or well the equation $(2)$ or well the equation $(3)$? In other case, if it is possible, explain why one should expect or well should not expect odd solutions for each one or some of the previous equations $(1)$, $(2)$ or $(3)$. I understand in this ocassion succinct answers, with the purpose to accept an useful answer. Many thanks.

I got computational evidence about the following conjecture based on my computation, but due that the integers in previous equations are exponentials I prefer don't add it. In any case I write the conjecture that I've evoked in the second paragraph.

Conjecture. There aren't odd integers $1\leq X$ satisfying $(1)$ and there aren't odd integers $1\leq X$ satisfying $(2)$. Similarly, an odd integer $X\geq 1$ for which $\varphi\left(X(2X+1)\right)\mid\varphi\left(X^{\sigma(X)}\sigma(X)^X\right)$ will never satisfy the identity $(3)$.

I think that, now, previous Conjecture it is just a speculation, thus if you can add heuristics or claims yourself computational evidence add it as comments. Also I hope not to bore you with this equations, therefore if you have ideas to improve the mathematical content this type of equations add a comment

  • $\begingroup$ Hi, all I think that an interesting question could be study $\operatorname{rad}(x)$ for $x$ a quasiperfect number. I can only to get obvious statements, if isn't in the literature feel free to study the question as a professional mathematician. $\endgroup$ – user142929 Aug 17 '19 at 10:37
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    $\begingroup$ Hi, @user142929. In case you still do not know, V. Siva Rama Prasad and C. Sunitha proved in (On the prime factors of a quasiperfect number, Theorem 3.5, page 18) that $$\operatorname{rad}(x) > \bigg(\frac{1}{2^{1/r} - 1}\bigg)^r$$ where $r = \omega(x)$ is the number of distinct prime factors of $x$. This is in response to your comment above about studying $\operatorname{rad}(x)$ for $x$ a quasiperfect number. $\endgroup$ – Arnie Bebita-Dris Jan 19 at 23:21
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    $\begingroup$ Using the lower bound in my previous comment together with the bound $r = \omega(x) \geq 7$ (Cohen G. L. & Hagis Jr., P. (1982). Some results concerning quasiperfect numbers, J. Austral. Math. Soc. (Ser.A), 33, 275–286), we obtain $$\operatorname{rad}(x) > \bigg(\frac{1}{2^{1/7} - 1}\bigg)^7 \approx 7553550.6198,$$ which implies that $\operatorname{rad}(x) \geq 7553551$ since $\operatorname{rad}(x)$ must be an integer. $\endgroup$ – Arnie Bebita-Dris Jan 19 at 23:31
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    $\begingroup$ Thanks you very much @JoseArnaldoBebita-Dris $\endgroup$ – user142929 Jan 20 at 8:41

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