# A conjecture concerning the equation $\sigma\left(\square\right)=\text{prime}$

I can deduce the following simple proposition, the definitions for $$\sigma(x)$$ the sum of divisors functions and $$\varphi(x)$$ the Euler totient function are assumed. After I present a conjecture that I've tested for the square $$1000\times 1000$$. (Please add feedback in comments if you think that the question can be improved before downvote it.) The sequence from the OEIS related to the post is A023194.

Claim. Let $$B\geq1$$ and $$C\geq 1$$ be positive integers with $$B$$ a prime numbers satisfying $$\sigma(C)=B,\tag{1}\label{1}$$ then $$\varphi(B)$$ has the factorization $$\varphi(B)=\operatorname{rad}(C)\cdot (B-C),\tag{2}\label{2}$$ where $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p$$ denotes the product of distinct primes dividing an integer $$n> 1$$.

One also can deduce under the same assumptions of Claim the following identities $$\varphi(C)=C\left(1-\frac{1}{\operatorname{rad}(C)}\right)$$, $$\frac{\varphi(C)}{C}=\frac{C-1}{B-1}$$ and $$\sigma(C^{1+\lambda})=\frac{C^{\lambda}-1}{\operatorname{rad}(C)-1}+BC^{\lambda}$$, for $$\lambda\geq 1$$ integer.

Conjecture. If \eqref{1} and \eqref{2} hold for some integers $$B$$ and $$C$$ strictly greater than $$1$$, then $$B$$ is a prime number.

Question. I would like to know if it is possible to prove previous conjecture, or what work can be done about it.

I was inspired in an edited question (on Mathematics Stack Exchange, now closed with identificator 4423186) that I'm going to delete from my profile on MSE.

As application (that I evoke) I tried, and I invite to it if you think that it is interesting, to combine these identities for an odd perfect number of the form (for example) $$BCM^2$$, for pairwise coprime integers $$B$$, $$C$$ and $$M$$ under the previous assumptions.

• I hope that there aren't typos in the simple deductions that I present here. Commented Jun 11, 2022 at 17:37
• MathJax note: \label+\eqref works here. Further, there is no need to end and begin italics around math mode; for example, "If \eqref{1} and \eqref{2} hold for some integers $B$ and $C$" *If \eqref{1} and \eqref{2} hold for some integers $B$ and $C$* works just fine, and you need not do "If \eqref{1} and \eqref{2} hold for some integers $B$ and $C$" *If \eqref{1} and \eqref{2} hold for some integers* $B$ *and* $C$. I have edited accordingly. Commented Jun 11, 2022 at 17:45
• Many thanks for your edit and remarks professor @LSpice Commented Jun 11, 2022 at 17:54
• As aside remark that isn't related to our post, is that at my home I define the arithmetic function $<x,y>=\varphi(y)-\operatorname{rad}(x)\cdot (y-x)$ for positive integers $x,y\geq 1$ and I call it the refraction bracket. This arithmetic function satisfies the properties a) $<x,x>=\varphi(x)$ $\forall x\geq 1$ integer and b) for a given integer $x$ and a prime number $y$ satisfying $\sigma(x)=y$ then $<x,y>=0$. While that at my home I call the chiral bracket the arithmetic funciton $[x,y]$ defined in our question of this MathOverflow with identificator 419954 Commented Jun 11, 2022 at 18:03
• Re, TeX note: please use $\langle x, y\rangle$ \langle x, y\rangle, not $<x, y>$ <x, y>. Notice especially the difference in spacing between, e.g., $2\langle x, y\rangle$ 2\langle x, y\rangle and $2<x, y>$ 2<x, y>. Commented Jun 11, 2022 at 18:32

Your conjecture is true. Notice first that it is enough to show that $$C=p^k$$ for some prime $$p$$. Indeed, in this case $$B=\sigma(C)=\frac{p^{k+1}-1}{p-1},$$ hence $$\mathrm{rad}(C)(B-C)=p\frac{p^k-1}{p-1}=B-1.$$ This means that $$\varphi(B)=B-1$$, so $$B$$ is prime.

To show that $$C$$ must be a prime power, assume the contrary. Let $$p$$ be the least prime factor of $$C$$. Since $$C$$ has at least one larger prime factor (otherwise $$C=p^k$$) , we have $$\mathrm{rad}(C)\geq p(p+1)=p^2+p.$$ On the other hand, $$B\geq \varphi(B)=\mathrm{rad}(C)(B-C)\geq (p^2+p)(B-C).$$ Therefore, $$C\geq B\left(1-\frac{1}{p^2+p}\right).$$ Next, if $$C=p_1^{k_1}\ldots p_l^{k_l}$$, then $$\sigma(C)=\prod_i \frac{p_i^{k_i+1}-1}{p_i-1}$$ and $$\frac{\sigma(C)}{C}=\prod_i\frac{p_i-p_i^{-k_i}}{p_i-1}.$$ Every factor in the product above is $$\geq 1$$, so we can bound the product from below by any one factor. In particular, for some $$k\geq 1$$ $$\frac{\sigma(C)}{C}\geq \frac{p-p^{-k}}{p-1}.$$ So $$B=\sigma(C)\geq C\frac{p-1/p}{p-1}=C\frac{p+1}{p}.$$ Substituting into the previous inequality, we obtain $$C\geq B\left(1-\frac{1}{p^2+p}\right)\geq C\left(1-\frac{1}{p^2+p}\right)\frac{p+1}{p}.$$ Dividing by $$C$$, we arrive at $$1\geq \left(1-\frac{1}{p^2+p}\right)\frac{p+1}{p}=1+\frac{1}{p}-\frac{1}{p^2}>1,$$ which is a contradiction.

• Many thanks for your excellent answer, as soon I can I read the nice details and I think that I'm going to understand the full proof, while that now I'm going to accept the answer. It's an honor for the site that persons and professional mathematicians as you are members of the site. Commented Jun 12, 2022 at 14:08