gpt pro 5.4 attempt on Math-5896

Problem 786

1) ROUND-2 OBJECTIVE

Path (C): identify the fatal obstruction in Round 1 and correct the statement. In Round 1 I treated the repetitions-allowed version as still open. That is incorrect. The currently curated literature record shows that, when repetitions are allowed, the problem is already settled negatively: every such set is contained in a level set of a real-valued additive function, and Erdős--Ruzsa--Sárközy imply an absolute upper-density barrier. The minimal corrected open problem is the distinct-elements version.

2) Round-1 FOUNDATION USED

I use the following Round-1 results without reproving them.

1. The finite additive-functional criterion: for finite $A$, the repetitions-allowed property is equivalent to the existence of a rational additive functional on prime-exponent vectors taking the value $1$ on every $a \in A$.
2. The Round-1 diagnosis that the real gap was to control level sets of additive functions.
3. The standard examples (the class $n \equiv 2 \pmod 4$, Selfridge's construction, and the one-large-prime finite construction).

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is an infinite-set extension of the Round-1 structural lemma. Instead of working only with finitely many primes and rational coefficients, I pass to the rational vector space of finitely supported prime-exponent vectors and extend a linear functional from the span of $\{\mathbf v(a): a \in A\}$ to the whole space. This gives a real-valued additive function on all natural numbers.

The second new ingredient is the corrected literature input: the updated 2026 discussion of Problem 786 records that Erdős--Ruzsa--Sárközy already settle the repetitions version negatively, and that the remaining open variant is the distinct-elements interpretation.

4) ATTACK PLAN (ROUND-2)

The exact gap left by Round 1 was this: the finite linear-algebra lemma had not yet been upgraded to arbitrary $A$, so it could not be combined with the known distribution theory of additive functions. I will:

1. extend the structural lemma from finite $A$ to arbitrary $A$;
2. reduce the repetitions-allowed problem to a level set $\{n : f(n)=1\}$ of a real additive function $f$;
3. import the Erdős--Ruzsa--Sárközy consequence recorded in the current literature summary, obtaining a negative answer to both the density and finite problems;
4. isolate precisely why this argument does not resolve the distinct-elements variant.

5) WORK (ROUND-2)

Write the repetitions-allowed property as follows: \[ a_1 \cdots a_r = b_1 \cdots b_s \quad (a_i,b_j \in A) \implies r=s. \]

Lemma 786.1 (global additive-functional criterion).

Let $A$ be a subset of the natural numbers with the repetitions-allowed property. Then there exists a real-valued additive function $f$ on the natural numbers such that \[ f(a)=1 \qquad \text{for every } a \in A. \]

Proof.

Let $P$ be the set of primes dividing at least one element of $A$. Let $V$ be the vector space over the rationals consisting of all finitely supported tuples \[ x=(x_p)_{p \in P}, \qquad x_p \in \mathbf Q. \] For a natural number $n$, let \[ \mathbf v(n) := (v_p(n))_{p \in P} \in V, \] where $v_p(n)$ is the exponent of $p$ in the factorisation of $n$.

Let $W$ be the rational span of $\{\mathbf v(a): a \in A\}$ inside $V$. Define a map $L$ on $W$ by declaring \[ L\!\left(\sum_{i=1}^m q_i \mathbf v(a_i)\right) := \sum_{i=1}^m q_i \] for rational numbers $q_i$ and elements $a_i \in A$.

I claim this is well-defined. Suppose \[ \sum_{i=1}^m q_i \mathbf v(a_i)=0. \] Choose a common denominator $D$ and set $m_i = D q_i \in \mathbf Z$. Then \[ \sum_{i=1}^m m_i \mathbf v(a_i)=0. \] Write $m_i = m_i^+ - m_i^-$ with $m_i^+,m_i^- \ge 0$. The equality of exponent vectors is equivalent to \[ \prod_{i=1}^m a_i^{m_i^+} = \prod_{i=1}^m a_i^{m_i^-}. \] Each side is a product of elements of $A$, with repetitions allowed. By the defining property of $A$, the total number of factors on the two sides must be equal: \[ \sum_{i=1}^m m_i^+ = \sum_{i=1}^m m_i^-. \] Hence \[ \sum_{i=1}^m m_i = 0, \] and dividing by $D$ gives $\sum_i q_i = 0$. So $L$ is well-defined.

Now extend $L$ from the subspace $W$ to a linear functional on all of $V$ by choosing a basis of $W$ and extending it to a basis of $V$. Keep the same name $L$ for the extension. Define \[ f(n) := L(\mathbf v(n)). \] Because \[ \mathbf v(mn)=\mathbf v(m)+\mathbf v(n), \] we have \[ f(mn)=f(m)+f(n) \] for all natural numbers $m,n$; that is, $f$ is additive. And for every $a \in A$, \[ f(a)=L(\mathbf v(a))=1. \] This proves the lemma. $\square$

Corollary 786.2.

Every repetitions-allowed set $A$ is contained in a level set of a real-valued additive function: \[ A \subseteq \{n : f(n)=1\} \] for some additive $f$.

External input.

The current 2026 update to the Erdős Problems page for Problem 786 records the following consequence of Erdős--Ruzsa--Sárközy $1973$: once a set is contained in a fixed level set of a real-valued additive function, one obtains \[ \overline d(A) \le \frac12, \] and, in the finite problem, there is an absolute constant $c>0$ such that for all sufficiently large $N$, \[ |A \cap [1,N]| \le (1-c)N. \] Applying this to Corollary 786.2 gives the promised negative answers.

Theorem 786.3 (repetitions version is already settled).

For the repetitions-allowed interpretation of Problem 786:

1. there is no set $A$ of density $>1-\varepsilon$ for every $\varepsilon>0$; in fact every such $A$ has upper density at most $1/2$;
2. in the finite problem one cannot have $|A|=(1-o(1))N$; indeed $|A| \le (1-c)N$ for some absolute $c>0$ and all sufficiently large $N$.

Why the distinct-elements version is genuinely different.

The proof above uses arbitrary integer relations among exponent vectors. Such a relation becomes a product identity only because repetitions are allowed. If one insists that all $a_i$ and all $b_j$ be distinct, then only disjoint $0$--$1$ relations are relevant; integer coefficients larger than $1$ are no longer legal. So the additive-function reduction collapses exactly at this point. This isolates the minimal corrected open statement: require the products on both sides to use distinct elements.

6) ADVERSARIAL VERIFICATION

1. The definition of $L$ uses only finitely many elements of $A$ at a time, so no infinite product or convergence issue appears.
2. The extension step is safe because $W$ is a rational vector subspace of $V$; I am extending a linear functional on vector spaces, not claiming that a homomorphism to the integers extends on a free abelian group. This avoids the standard obstruction exemplified by $2\mathbf Z \to \mathbf Z$.
3. The conclusion is only $A \subseteq \{f=1\}$, not equality; this is the correct statement needed for the upper-bound transfer.
4. The distinct-elements obstruction is real: the argument from an integer relation \[ \sum m_a \mathbf v(a)=0 \] produces repeated factors $a^{m_a^+}$ and $a^{m_a^-}$, which are illegal in the distinct-elements variant whenever some coefficient has absolute value bigger than $1$.

7) FINAL

FULL SOLUTION -- COUNTEREXAMPLE / DISPROOF.

For the repetitions-allowed interpretation, both optimistic conclusions are false. Every such set is contained in a level set of a real-valued additive function, and the Erdős--Ruzsa--Sárközy consequences recorded in the current literature give upper density at most $1/2$ and a finite upper bound $|A| \le (1-c)N$. The minimal corrected open problem is the distinct-elements version.

8) COMPLETION ESTIMATE

COMPLETION: 100%

9) REFERENCES

References

• [ERS73-786] P. Erdős, I. Ruzsa, and A. Sárközy, On the number of solutions of $f(n)=a$ for additive functions, Acta Arith. 24 $1973$, 1--9.
• [Bloom786] T. F. Bloom, Erdős Problem #786, updated 2 February 2026, https://www.erdosproblems.com/786.
• [Bloom786Thread] T. F. Bloom, T. Tao, and others, Discussion thread for Erdős Problem #786, especially comments of 18 October 2025 and 2 February 2026, https://www.erdosproblems.com/forum/thread/786.