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**](https://math.stackexchange.com/questions/4423186/on-the-equation-sigma-left-square-right-textprime-propositions-that-can)) 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.



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


