GPT 6 Astra Ultra attempt on Math-5441
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Contents
FORMAL RESTATEMENT
Erdős #121; partial reconstruction, 2026-09-23. Let $F_k(N)$ be the maximum size of $A\subseteq[N]$ for which no $k$ distinct elements have square product. Is $F_5(N)=(1-o(1))N$, and more generally does this hold for odd $k\ge5$?QUICK LITERATURE AND CONTEXT CHECK
The assertion is known to be false: Tao's primary paper proves $F_k(N)\le(1-c_k)N$ for sufficiently large $N$ and fixed $k\ge4$, with $c_k>0$. We prove elementary comparison results and the precise probabilistic transfer step, while isolating the analytic construction needed for the full negative answer. The latter is cited, not claimed as independently proved here. No novelty is claimed.WORK
The two-element case
Write uniquely $n=s(n)t^2$ with $s(n)$ squarefree. Then $uv$ is a square if and only if $s(u)=s(v)$. An admissible set for $F_2$ contains at most one number per kernel, and the squarefree numbers themselves achieve this bound. Thus \[ F_2(N)=\sum_{n\le N}\mu(n)^2 =\sum_{d\le\sqrt N}\mu(d)\lfloor N/d^2\rfloor =\frac{6}{\pi^2}N+O(\sqrt N). \] The final equality uses the absolutely convergent Euler product $\sum_d\mu(d)/d^2=1/\zeta(2)$; its tail and the floor errors are each $O(\sqrt N)$.A large set avoiding every odd square product
Let $A_N$ consist of integers at most $N$ with a prime divisor $p>\sqrt N$. Each has exactly one such prime, appearing to exponent one. In a product of an odd number of members, the total number of these large-prime occurrences is odd, so at least one prime occurs an odd number of times. The product cannot be a square, whether or not the selected elements are distinct. Consequently \[ F_k(N)\ge |A_N|\quad\hbox{for every odd }k. \] Distinct primes above $\sqrt N$ cannot divide the same integer at most $N$, so, by the prime number theorem and Mertens' prime reciprocal sum, \[ |A_N|=\sum_{\sqrt N<p\le N}\lfloor N/p\rfloor =N\sum_{\sqrt N<p\le N}\frac1p+O(\pi(N)) =(\log2+o(1))N. \tag{1} \] This is a genuine positive-density lower construction, but does not approach density one.A transfer lemma isolating the upper-bound difficulty
Suppose a random tuple $(X_1,\ldots,X_k)$ has square product always, and there is an event $E$ such that $X_i\in[N]$ on $E$ and \[ \Pr(E)\ge aL^{-k},\qquad \Pr(E\cap\{X_i=x\})\le b/(NL^k),\qquad \Pr(\exists i<j:X_i=X_j)=o(L^{-k}), \tag{2} \] where $L=\log N$ and $a,b>0$ depend only on $k$. If $A\subseteq[N]$ omits at most $aN/(4kb)$ elements, the union bound shows that the probability of $E$ together with some $X_i\notin A$ is at most $a/(4L^k)$. For sufficiently large $N$, the collision probability is also at most $a/(4L^k)$. Thus with positive probability all the $X_i$ are distinct members of $A$ and have square product. This proves \[ F_k(N)\le (1-a/(4kb))N \tag{3} \] under $2$.The algebraic source of such tuples is particularly simple. Give the edges of a complete graph on $k$ vertices independent positive integer labels $D_{ij}=D_{ji}$ and put $X_i=\prod_{j\ne i}D_{ij}$. Then $\prod_iX_i=\prod_{i<j}D_{ij}^2$ automatically. What is difficult is choosing the distributions so the tuple has the uniform marginal bounds in $2$ after conditioning on its size. Uniform edge labels alone do not establish those bounds. Tao's Proposition 2.1 constructs precisely $2$, using small squarefree factors, larger prime factors, and prime-distribution estimates. Thus $3$, together with that external proposition, reproduces the known negative answer. Its uniform probability estimate has not been proved in this attempt.