gpt 5.2 attempt on Math-658

OUTPUT 1) FORMAL RESTATEMENT Definitions/Conventions. - $\phi(n)$ is Euler's totient function. - For $a\in\mathbb N$, say $a$ is a totient if $\exists n\in\mathbb N$ with $\phi(n)=a$. - For a totient $a$, define \[n_a:=\min\{n\in\mathbb N:\phi(n)=a\}.\]

Question. Does there exist an infinite set $A\subset\mathbb N$ of totients such that \[\lim_{a\to\infty,\ a\in A}\frac{n_a}{a}=+\infty?\]

2) QUICK LITERATURE/CONTEXT CHECK Web browsing is not available in this session. From the problem text only: - Carmichael asked whether $\phi(n)=t$ can have exactly one solution. - Erdős proved that if such a $t$ exists then infinitely many such $t$ exist. I do not use these claims as lemmas below.

3) ATTACK PLAN Proof track ideas (existence of such an $A$): 1. Try to force any solution $n$ of $\phi(n)=a$ to have many distinct small prime factors, making $n/a=n/\phi(n)=\prod_{p\mid n}p/(p-1)$ very large. 2. Pick $a$ with carefully engineered divisibility constraints so that any preimage must include certain primes (e.g. by forcing many primes $p$ with $p-1\mid a$), and then argue that a smaller preimage cannot exist.

Disproof track ideas (nonexistence): 1. Try to show a universal upper bound $n_a \le C a (\log\log a)^{O(1)}$ or even $n_a\le Ca$ for all totients $a$. 2. Look for mechanisms producing small preimages, e.g. primes $p$ with $p-1=a$ or other constructions.

Chosen path in this attempt: prove two unconditional lemmas about the shape of totients/minimal preimages and run a sanity-check computation of $n_a/a$ for $a$ up to $3\times 10^5$.

4) WORK Lemma 1 (Parity obstruction: $\phi(n)$ is even for $n>2$). If $n>2$ then $\phi(n)$ is even. In particular, every totient $a>2$ is even.

Proof. If $n$ is divisible by an odd prime $p$, then $\phi(n)$ is divisible by $\phi(p)=p-1$, which is even, so $\phi(n)$ is even. If $n$ has no odd prime factor, then $n=2^k$ with $k\ge 2$ (since $n>2$), and $\phi(2^k)=2^{k-1}$ is even. Thus in all cases $n>2\Rightarrow \phi(n)$ even. $\square$

Lemma 2 (Prime totients have tiny minimal preimage). Let $p$ be an odd prime and set $a:=p-1$. Then $a$ is a totient (since $\phi(p)=p-1$), and the minimal preimage equals $n_a=p$. Consequently, \[\frac{n_a}{a}=\frac{p}{p-1}\to 1\quad\text{as }p\to\infty.\]

Proof. We have $\phi(p)=p-1=a$, so $n_a\le p$. On the other hand, for any $n>1$, $\phi(n)\le n-1$ (since among $1,\dots,n$ only $n$ itself is not counted, and at least one number is not coprime unless $n$ is prime; the inequality $\phi(n)\le n-1$ is always true). Thus if $\phi(n)=a=p-1$, then $n-1\ge \phi(n)=p-1$, so $n\ge p$. Therefore $n_a=p$. The ratio statement is immediate. $\square$

Lemma 3 (An infinite family with uniformly bounded $n_a/a$). Let $m\ge 1$ and set $a:=2^m$. Then $a$ is a totient because $\phi(2^{m+1})=2^m=a$, and hence \[\frac{n_a}{a}\le \frac{2^{m+1}}{2^m}=2.\]

Proof. Compute $\phi(2^{m+1})=2^{m+1}-2^m=2^m=a$. By definition of $n_a$ as the smallest preimage, $n_a\le 2^{m+1}$. Divide by $a=2^m$ to get $n_a/a\le 2$. $\square$

FAST REALITY CHECK (local computation; $a\le 300{,}000$). I computed $\phi(n)$ for $1\le n\le 300{,}000$ and for each totient value $a\le 300{,}000$ recorded the smallest $n_a\le 300{,}000$ with $\phi(n_a)=a$. Among these $a$, the largest ratios $n_a/a$ observed were about $2.0$: top examples (within this search bound): \[ \begin{array}{c|c|c} a & n_a & n_a/a\\\hline 5888 & 11985 & 2.035\\ 10496 & 21165 & 2.016\\ 17408 & 34935 & 2.007\\ 32768 & 65535 & 2.000 \end{array} \] Among $a\ge 50{,}000$ (still with $n_a\le 300{,}000$), the largest ratios found were $\approx 1.996$. This is consistent with (but does not prove) the possibility that $n_a/a$ might remain bounded, contrary to the question's desired conclusion.

5) VERIFICATION - Lemma 1 is standard and the proof splits into "odd prime factor" vs "power of two" cases. - Lemma 2 uses only the elementary inequality $\phi(n)\le n-1$ and the explicit solution $n=p$. - Lemma 3 is a direct computation. - Computation is limited to preimages $n\le 300{,}000$, so it cannot rule out larger minimal preimages for larger $a$.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: I proved basic obstructions and easy families: every totient $a>2$ is even (Lemma 1), totients of the form $a=p-1$ (odd prime $p$) have $n_a/a\to 1$ (Lemma 2), and the infinite family $a=2^m$ has $n_a/a\le 2$ (Lemma 3).

(ii) First gap: Construct an infinite set $A$ of totients for which every solution $n$ to $\phi(n)=a$ is forced to have very large $n/a$, or prove a general upper bound on $n_a/a$ that rules out divergence.

(iii) Top 3 next moves (concrete): 1. Try to force many primes into any preimage: choose $a$ so that for many primes $p$ one has $p-1\mid a$, and show that any $n$ with $\phi(n)=a$ must be divisible by many of those $p$. 2. Study the structure of minimal preimages: prove lemmas of the form "if $\phi(n)=a$ then there exists a solution $n'\le C(a)$ with specified constraints", aiming to bound $n_a$. 3. Extend computation to much larger bounds (e.g. $n\le 10^8$) and track the growth of $\max_{a\le X} n_a/a$; if it stays bounded, attempt to conjecture a sharp constant.

(iv) What a minimal counterexample would likely look like (structure/parameters): An infinite sequence of totients $a_k$ such that every $n$ with $\phi(n)=a_k$ must include an increasing set of small primes in its factorization, forcing $n/\phi(n)=\prod_{p\mid n}p/(p-1)\to\infty$, while simultaneously preventing the existence of alternative preimages with fewer primes (and hence smaller $n$).