GPT 6 Astra Ultra attempt on Math-5483
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Erdős Problem 186: the growth of non-averaging sets
1) Statement and scope
A set $A$ of integers is non-averaging if no member is the average of a subset of at least two other, distinct members. Let $F(N)$ be the maximum size of such a subset of $[N]$. The known exponent is \[ F(N)=N^{1/4+o(1)}. \] We reconstruct the elementary lower construction and combine it with the sharp exponent upper bound of Pham and Zakharov, used as an external theorem. This does not assert the stronger estimate $F(N)=\Theta(N^{1/4})$.2) A quadratic construction with a large linear term
Fix an integer $q\geq2$ and put \[ a_i=iq^3+\frac{i(i+1)}2\quad(1\leq i\leq q-1), \qquad A_q=\{a_1,\ldots,a_{q-1}\}. \] These are positive, distinct integers, and \[ a_{q-1}=(q-1)q^3+q(q-1)/2<q^4. \] Suppose $a_i$ is the average of a subset indexed by $S\subseteq\{1,\ldots,q-1\}\setminus\{i\}$, where $k=|S|\geq2$. Multiplying the average equation by $k$ gives \[ q^3\left(\sum_{j\in S}j-ki\right) +\sum_{j\in S}\frac{j(j+1)}2-k\frac{i(i+1)}2=0. \tag{1} \] Every quadratic term lies between zero and $q(q-1)/2$. The absolute value of the second part of $1$ is at most $kq(q-1)/2<q^3$, because $k\leq q-1$. The first part is an integer multiple of $q^3$. Both parts must therefore be zero. It follows that \[ \sum_{j\in S}j=ki,\qquad \sum_{j\in S}j^2=ki^2. \] Consequently \[ \sum_{j\in S}(j-i)^2 =ki^2-2i(ki)+ki^2=0. \] Each summand is nonnegative, so every $j$ in $S$ equals $i$, contrary to the choice of $S$. Thus $A_q$ is non-averaging.For arbitrary large $N$, take $q=\lfloor N^{1/4}\rfloor$. Then $A_q\subseteq[N]$ and \[ F(N)\geq q-1\gg N^{1/4}. \tag{2} \] This proves the lower bound for all large $N$, not just fourth powers. The construction is of the classical Bosznay type; no novelty is claimed.
3) The external upper theorem and the exponent
Pham--Zakharov prove that, for every $\epsilon>0$, there is an $N_0(\epsilon)$ such that every non-averaging subset of $[N]$ has size at most $N^{1/4+\epsilon}$ when $N\geq N_0(\epsilon)$. This structure-theoretic upper bound is the external input. Combining it with $2$, taking logarithms, and then letting $\epsilon\downarrow0$ yields \[ \lim_{N\to\infty}\frac{\log F(N)}{\log N}=\frac14, \] which is the displayed exponent asymptotic. The definition using a subset not containing its average agrees with the stated one: a singleton of a different element cannot have that average, while a subset containing the average can have it removed without changing the equality.4) Outcome and sources
Outcome: SOLVED at the known exponent precision, using the explicit external upper theorem. The lower construction and limiting deduction are proved here. The Pham--Zakharov structural argument is not independently reproved.- H. T. Pham and D. Zakharov, Sharp bound for the Erdős--Straus non-averaging set problem,
https://arxiv.org/abs/2410.14624. - T. F. Bloom, Problem 186, including the attribution of the lower bound to Bosznay,
https://www.erdosproblems.com/186, checked 23 September 2026.