In this post we denote the Euler's totient function that counts the number of positive integers $1\leq k\leq n$ such that $\gcd(k,n)=1$ as $\varphi(n)$, and the sum of divisors function $\sum_{1\leq d\mid n}d$ as $\sigma(n)$. As reference I add the Wikipedia Sophie Germain prime that provides the well-known definition that a prime $p$ is a Sophie Germain prime if $2p + 1$ is also prime. This prime constellation corresponds to the entry A005384 from the OEIS.
From previous definition and the calculations of previous arithmetic function $\varphi(n)$ and $\sigma(n)$ it is easy to check the proof of the following claim.
Claim. If $x=p$ is a prime such that $y=2p+1$ is also prime, then $(x,y)$ solves the equation $$\sigma(x^{\varphi(y)})=\frac{1}{\varphi(x)}(x^y-1).\tag{1}$$
We propose the following conjecture inspired in previous claim, from the substitution $$y=x+\sigma(x).\tag{2}$$
Conjecture. Let $S\geq 1$ be an integer that satisfies $$1+\varphi(S)\sigma(S^{\varphi(S+\sigma(S))})=S^{S+\sigma(S)},\tag{3}$$
then $S$ is a Sophie Germain prime.
The motivation of this question is try to get a characterization of this prime constellation, Sophie Germain primes, in terms of the important number theoretic functions $\sigma(n)$ and $\varphi(n)$ encoded in an equation relating those. Are known the importance of Sophie Germain primes, and on the other hand also inequalities involving the mentioned number theoretic functions.
Question. What work can be done with the purpose to prove or refute previous Conjecture? Many thanks.
Computational evidence. You can check in the web Sage Cell Server this line written in Pari/GP
for(x=1, 1559, if(1+eulerphi(x)*sigma(x^eulerphi(x+sigma(x)))==x^(x+sigma(x)),print(x)))
just copy and paste it to evaluate in the web choosing as Language the option GP.
The following, and similar scripts that I can to evaluate for(x=1, 5000, if((isprime(x)==0||isprime(2*x+1)==0)&&1+eulerphi(x)*sigma(x^eulerphi(x+sigma(x)))==x^(x+sigma(x)),print(x)))don't show outputs as counterexamples.
Best Answer
Nice question! Assume that $p^2 \mid S$. This would then give $p \mid \varphi(S)$. However, this is impossible in our case since: $$p \mid S^{S+\sigma(S)}-\varphi(S)\sigma(S^{\varphi(S+\sigma(S))}) \implies p \mid 1$$ which is clearly false.
Since $S$ is square-free, let: $$S=\prod_{i=1}^n p_i$$ for distinct primes $p_1,p_2, \ldots ,p_n$. Then, by substitution: $$1+\varphi(\prod p_i)\sigma(\prod p_i^{\varphi(S+\sigma(S))})=\prod p_i^{S+\sigma(S)}$$
$$1+\bigg(\prod (p_i-1)\bigg)\bigg(\prod \frac{p_i^{\varphi(S+\sigma(S))+1}-1}{p_i-1}\bigg)=\prod p_i^{S+\sigma(S)}$$
$$\bigg(\prod (p_i^{\varphi(S+\sigma(S))+1}-1)\bigg)=\bigg(\prod p_i^{S+\sigma(S)}\bigg)-1$$
But, we have:
$$\bigg(\prod (p_i^{\varphi(S+\sigma(S))+1}-1)\bigg) \leqslant \bigg(\prod (p_i^{S+\sigma(S)}-1)\bigg) \leqslant \bigg(\prod p_i^{S+\sigma(S)}\bigg)-1$$
We need equality in both places. The first equality can only hold true when $S+\sigma(S)$ is prime, so that $\varphi(S+\sigma(S))+1=S+\sigma(S)$. The second inequality can hold true only when there is one term in the product, i.e. $n=1$.
Since $N=1$, $S$ is prime. This shows that $S+\sigma(S)=2S+1$ is prime. Hence, $S$ is a Sophie-Germain prime, proving your conjecture to be true!