Erdős Problem #1194
Status Open · erdosproblems.comLLM status Open unverified
Erdős problems openaigpt 6 astra ultra 7% cataloguedOpenOpenStatement
Suppose $A\subseteq\mathbb N$ has exactly one ordered pair $(a_n,b_n)\in A^2$ satisfying $a_n-b_n=n$ for every positive integer $n$. Determine the growth that the ratios $a_n/n$ must exhibit, and how slowly they can grow across such perfect difference sets.
Restatement from Math-6248-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Math-6248
Erdős Problem 1194 --- GPT 6 Astra Ultra partial attempt
FORMAL RESTATEMENT
Suppose $A\subseteq\mathbb N$ has exactly one ordered pair $(a_n,b_n)\in A^2$ satisfying $a_n-b_n=n$ for every positive integer $n$. Determine the growth that the ratios $a_n/n$ must exhibit, and how slowly they can grow across such perfect difference sets.CONTEXT AND SCOPE
This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.Here a_n is the larger member of the pair representing the difference n; it is not the nth element of A. The source is an open-ended growth question, not an assertion that the ratios increase monotonically.
ATTACK PLAN AND WORK
There is a precise transfer from sparse truncations to large representing endpoints. Let $t=A(X)$. At most $\binom t2$ positive differences can be represented with larger endpoint at most $X$, because every such pair lies in $A\cap[1,X]$ and all positive differences are unique. Put $M=\binom t2+1$. Among the differences $1,\ldots,M$, at least one therefore has \[a_n>X,\qquad n\le M,\qquad \frac{a_n}{n}>\frac X{\binom{A(X)}2+1}.\tag{1}\] This is an exact inequality, not an assertion that the function $a_n/n$ is monotone.For example, if along some $X_j\to\infty$ one has $A(X_j)\le K\sqrt{X_j/\log X_j}$, then $1$ forces $a_{n_j}/n_j\gg_K\log X_j\ge\log n_j$ up to a fixed constant for large $j$. The indices $n_j$ must be unbounded, since each of finitely many endpoints $a_n$ is fixed while $X_j$ diverges. This proves the deduction from a logarithmic sparsity input; it does not independently prove that input.
The official page describes such sparsity results and substantially stronger subsequent endpoint bounds. They are context, not achievements of this short argument. Equation $1$ identifies exactly which counting estimate would have to be strengthened to improve this particular route.