The Family $X_{qp^s}$ for Distinct Primes $q,p$ in Connection with Carmichael's Totient Conjecture

Source abstract

Shannon, Shiue, Saito and the author \cite{SSHS2025} studied the sets $X_k=\{n\in\mathbb{N}:\varphi(n)=k\}$ in connection with Carmichael's totient conjecture, and proved that for an odd prime $p\neq3$ and $s\in\{1,2,3,4\}$, $X_{2p^s}$ is empty whenever $2p^s+1$ is not prime (with $s=1$ fully characterized in both directions), conjecturing that this persists for every $s\ge1$. We resolve that conjecture in full, obtaining a complete characterization of $X_{2p^s}$ for every $s\ge1$ and every odd prime $p\neq3$, together with a parity dichotomy: $X_{2p^s}=\varnothing$ unconditionally whenever $s$ is even, while for odd $s$ both outcomes genuinely occur, determined exactly by the primality of $2p^s+1$. We then close the one remaining gap at $p=3$, where the proof technique degenerates, with a separate, unconditional characterization of $X_{2\cdot3^s}$. Building on these results, we consider the natural generalization $X_{qp^s}$, where $q\neq p$ are distinct primes and $s\ge1$. We show that this generalization splits into three cases of very different character. If $q$ and $p$ are both odd, the problem is trivial: $qp^s$ is odd, so $X_{qp^s}=\varnothing$ by elementary parity, with no case-analysis needed. If $q=2$, the problem is exactly the family $X_{2p^s}$ resolved above. If $p=2$, giving the ``mirror'' family $X_{2^sq}$ for an odd prime $q$, the problem is genuinely different. Beyond $s=1$, we give a complete resolution for $s=2$, recovering, with an explicit proof and solution sets, the qualitative dichotomy of \cite[Thm.~2.3.1]{SSHS2025}, but the problem remains open for $s\ge3$.