gpt pro 5.2 attempt on Math-5970

1) FORMAL RESTATEMENT

For each $n\ge 1$, let $f(n)$ be the maximum size of a set $A\subseteq\{1,2,\dots,n\}$ with the property:

For all $a,b,c,d\in A$ with $a\le b\le c\le d$, if $abcd$ is a perfect square, then $ad=bc$.

Ambiguity note. Because $A$ is a set, elements of $A$ are distinct, but the statement does not explicitly say whether the quadruple $(a,b,c,d)$ is required to consist of four distinct elements of $A$ or whether repetitions are allowed (e.g. $(a,b,b,c)$). The condition as written quantifies over $a,b,c,d\in A$ and allows equality in $a\le b\le c\le d$, so the literal reading allows repetitions. In what follows, I adopt this literal (stronger) reading; any set satisfying it also satisfies the "distinct quadruples only" variant.

The question asks for the asymptotic order of $f(n)$ (e.g. is $f(n)=o(n)$?).

2) QUICK LITERATURE/CONTEXT CHECK

The problem text attributes the question to Erdős--Sárközy--Sós and reports (without proof here) that Erdős claimed Sárközy proved $f(n)=o(n)$. The problem text also notes a lower bound $f(n)\gg n/\log n$ coming from taking $A$ to be the primes.

3) ATTACK PLAN

Proof-direction (upper bounds).

• Convert the condition "$abcd$ square" into parity constraints on prime exponents (squarefree kernels), turning the problem into forbidding certain 4-term relations; try to show this forces sparseness.
• Search for a combinatorial reformulation (hypergraph of forbidden 4-tuples) and apply container-style counting or energy bounds.

Disproof-direction (large constructions).

• Look for multiplicative constructions beyond primes (e.g. numbers with restricted squarefree part) that still avoid nontrivial square quadruples.
• Use computation for moderate $n$ to guess the true growth of $f(n)$ under the intended interpretation.

4) WORK

Lemma 888.1 (Primes give a valid set). Let $A$ be the set of primes $\le n$. Then $A$ satisfies the stated property (both in the "repetitions allowed" and "distinct only" interpretations). Consequently $f(n)\ge \pi(n)$.

Proof. Take $a\le b\le c\le d$ in $A$ (so each is prime) and assume $abcd$ is a perfect square. Write the prime factorization of $abcd$: since $a,b,c,d$ are primes, each of these primes occurs with exponent equal to its multiplicity among $\{a,b,c,d\}$. For $abcd$ to be a square, every prime exponent must be even. Hence every prime appearing among $a,b,c,d$ must appear an even number of times. Because there are four primes total, the multiset $\{a,b,c,d\}$ must be either:

• $\{p,p,p,p\}$ for some prime $p$, or
• $\{p,p,q,q\}$ for two distinct primes $p<q$.

Under the imposed ordering $a\le b\le c\le d$, the second case forces $a=b=p$ and $c=d=q$. In either case we have $ad=bc$ (both sides equal $p^2$ in the first case, and both equal $pq$ in the second). Thus the property holds for all such quadruples, and $|A|=\pi(n)$ gives $f(n)\ge \pi(n)$. $\square$

Lemma 888.2 (Square-product pairs force a geometric mean). Assume $A\subseteq\{1,\dots,n\}$ satisfies the stated property (with repetitions allowed). Let $a,c\in A$ with $a<c$ and suppose $ac$ is a perfect square. Then there is at most one element $b\in A$ with $a<b<c$. Moreover, if such a $b$ exists, then $b^2=ac$ (hence $b=\sqrt{ac}\in\mathbb{N}$).

Proof. Assume $a<c$ are in $A$ and $ac$ is a square. Let $b\in A$ satisfy $a<b<c$. Consider the nondecreasing quadruple $(a,b,b,c)$; it is allowed because repetitions are permitted and $a\le b\le b\le c$. Its product is \[ a\cdot b\cdot b\cdot c = b^2\,(ac). \] Because $b^2$ is a perfect square and $ac$ is assumed to be a perfect square, their product $b^2(ac)$ is a perfect square. Therefore the defining property of $A$ applies and yields \[ a\,c = b\,b = b^2. \] So any such $b$ must equal $\sqrt{ac}$. In particular, there is at most one element of $A$ strictly between $a$ and $c$. $\square$

Fast reality check (computation, literal "repetitions allowed" version). I brute-forced the maximum size $f(n)$ for $n\le 15$ by enumerating all subsets $A\subseteq\{1,\dots,n\}$ and checking the condition for all nondecreasing quadruples $a\le b\le c\le d$ with entries in $A$ (allowing repetitions). The exact maxima found were: \[ \begin{array}{c|ccccccccccccccc} n & 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15\\\hline f(n) & 1&2&3&3&4&5&6&6&6&7&8&8&9&10&10 \end{array} \] One maximizing example for $n=15$ is \[ A=\{1,2,3,5,6,7,10,11,13,14\},\qquad |A|=10, \] and it was verified directly (by exhaustive quadruple checking) that this $A$ has no violating quadruple.

5) VERIFICATION

Lemma 888.1: the proof uses only unique factorization and the characterization of square integers by even prime exponents.

Lemma 888.2: the key step is that $b^2(ac)$ is a square if and only if $ac$ is a square, because $b^2$ is always a square. Then the defining implication gives $ac=b^2$.

Computation: the brute check tested exactly the implication "if $abcd$ is a square then $ad=bc$" over all nondecreasing quadruples with repetition; this matches the literal interpretation.

6) FINAL

UNRESOLVED

1. (i) Strongest proved partial result here. The set of primes gives a valid construction of size $\pi(n)$ (Lemma 888.1), so $f(n)\ge \pi(n)\gg n/\log n$. Also, if $a,c\in A$ have $ac$ square, then any $b\in A$ with $a<b<c$ is forced to equal $\sqrt{ac}$ (Lemma 888.2).
2. (ii) First gap (crisp). Prove a nontrivial upper bound on $f(n)$ (ideally $f(n)=o(n)$, or any explicit $f(n)\le n/(\log n)^{\delta}$) from the square-quadruple implication.
3. (iii) Top 3 next moves.
   • Reformulate the constraint in terms of squarefree kernels: write each $x\in A$ as $x=s(x)\,t(x)^2$ with $s(x)$ squarefree, and analyze which 4-tuples of $s(x)$ can yield a square product without forcing $ad=bc$.
   • Push the computation to larger $n$ under both interpretations (repetitions allowed vs. distinct quadruples only) to guess the true growth rate and to detect structure in extremal sets.
   • Try to prove that many elements sharing the same squarefree kernel create many square quadruples; quantify this to obtain an upper bound by double counting.
4. (iv) Minimal counterexample structure. To refute $f(n)=o(n)$ one would need a sequence $n_t\to\infty$ and sets $A_t\subseteq\{1,\dots,n_t\}$ with $|A_t|\ge \delta n_t$ for some fixed $\delta>0$ such that every nondecreasing quadruple from $A_t$ with square product satisfies $ad=bc$. Such a construction would have to prevent the "parity cancellations" of prime exponents that normally generate many square products among dense sets.