I am currently trying to describe solutions to the equation
\begin{equation} \varphi(n)+s=\varphi(n+s) \end{equation}
where $s \equiv 2 \pmod{4}$, $n$ is even, and $\varphi(n)$ is the Euler totient function. This is part of a more general inquiry, but that's not especially pertinent to this specific question. The fact that $s \equiv 2 \pmod{4}$ is useful, since it forces one of $\varphi(n), \varphi(n+s)$ to be congruent to $2 \pmod{4}$ as well, implying that the argument (either $n$ or $n+s$, which are even) has only one odd prime factor. That is, if $\varphi(n) \equiv 2 \pmod{4}$, then $n=2p^k$ for some odd prime $p$, and the same goes for $n+s$.
Fascinatingly, after writing some simple python code to search for solutions to the equation up to 100,000 (I know, not very high at all, but I'm using the free version of sagemathcloud and the computer poops out if I go much higher), nearly all solutions $n$ have $n+s=2p$; i.e., it is practically always $n+s$ and never $n$ that is congruent to $2 \pmod{4}$, and $k=1$. Indeed, the only counterexamples to this for $s=2, 6, 14$ are very small; i.e., less than 100. A priori, I don't see any reason to suspect that it should be $n+s$ rather than $n$ that is equal to $2p^k$, and it certainly doesn't seem obvious why $k=1$.
If you have any thoughts, suspicions, or insights into this strange trend, I'd love to hear them. I've been working on this and similar problems for a while, and my thought processes are starting to repeat themselves, rather than coming up with anything new.