# Composite $n$ such that $\sigma(n) \equiv n+1 \pmod{\phi(n)}$

I'm looking for composite $n$ such that $$\sigma(n)\equiv n+1\pmod{\varphi(n)}$$

Are there only finitely many? Can this be proved?

This is Sloane's A070037 but there's not much information in the entry.

-

This isn't an answer, only the easy fact that I can deduce. There is an integer $k\geq 1$ such that $$\sigma(n)=k\phi(n)+n+1,$$ thus by exponentiation, where you take the basis as $A>1$, a positive integer such that $gcd(A,n)=1$ (for example, take the first prime that is greater than $n$), then by Euler-Fermat theorem $$A^{\sigma(n)}= \left( A^{k} \right)^{\phi(n)} A^{n+1} \equiv A^{n+1}\mod n,$$ thus there is a positive integer $K$ such that $$A^{\sigma(n)}=Kn+A^{n+1}.$$
Now on assumption that there are infinitely many of these $n_{\rho}$ solutions of the conguence in your problem (thus this is a hypothesis to work, looking to explore the problem or close it with a proof by contradiction) using Gronwall's Theorem, I can deduce (labelling and) taking the exponentiation $\frac{1}{n_{\rho}\log\log n_{\rho}}$
$$\limsup_{n_{\rho}\to\infty} \left( K_{\rho}n_{\rho}+A_{\rho}^{n_{\rho+1}} \right)^{\frac{1}{n_{\rho}\log\log n_{\rho}}}=\limsup_{n_{\rho}\to\infty}A_{\rho}^{\frac{\sigma(n_{\rho})}{n_{\rho}\log\log n_{\rho}}}=(\limsup_{n_{\rho}\to\infty}A_{\rho})^{e^\gamma},$$ where $\gamma$ is the Euler-Mascheroni constant. I don't know how take advantage from these, I don't know how to study lower bound for the sequences $K_{\rho}$ and $A_{\rho}$. Please say me when you consider if this contribution is good to appear as a partial answer in your good question. Thanks.
I hope that my computation in RHS for $\lim sup$ is right. – user243301 May 12 at 14:06