gpt pro 5.4 attempt on Math-5494

Erdős Problem #203

1) ROUND-2 OBJECTIVE

Path pursued: (C)-style obstruction sharpening / strict advance without a full resolution.

Round 1 proved a finite-box lemma: for any fixed finite set of exponent pairs $(k,\ell)$, one can choose $m$ so that all corresponding values $2^k3^\ell m+1$ are composite. The obvious gap was that this did not control infinitely many exponent pairs.

The most promising Round 2 move is therefore to prove a strictly stronger theorem that kills infinitely many exponent pairs at once in a structured way, and then identify the precise remaining obstruction.

The new goal in this round is to pass from "finite boxes" to "finite collections of infinite rays" in the exponent lattice.

2) Round-1 FOUNDATION USED

I use the following Round 1 items.

• The formal restatement of the problem.
• The Round 1 finite-box lemma: any finite set of exponent pairs $(k,\ell)$ can be forced composite simultaneously by a CRT construction.
• The Round 1 observation that for odd $m>1$, the value at $(k,\ell)=(0,0)$ is \[ m+1, \] which is even and at least $4$, hence automatically composite.

I do not reprove the finite-box lemma. It is used only as the comparison point that Round 2 must beat.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is a theorem of Filaseta on generalized Sierpiński numbers:

For every fixed integer $A\ge 2$, there exists an arithmetic progression of odd positive integers $m$ such that \[ m a^n+1 \] is composite for every integer $a$ with $2\le a\le A$ and every $n\ge 1$.

This is dramatically stronger than the Round 1 finite-box lemma. It controls infinitely many exponents $n$ simultaneously for every base $a$ in a fixed finite interval.

The exponent pairs $(k,\ell)$ for the original problem naturally decompose into rays: \[ (k,\ell)=n(u,v), \] where $(u,v)$ is primitive, meaning $\gcd(u,v)=1$, and $n\ge 1$.

Thus Filaseta's theorem lets us kill every pair on any fixed finite collection of primitive rays, because each ray corresponds to the fixed base \[ a=2^u3^v. \]

4) ATTACK PLAN (ROUND-2)

The exact Round 1 gap was: \[ \text{finite sets of exponent pairs are controllable, but infinite structured families were not.} \]

To overcome this, I will prove:

1. a finite-ray annihilation theorem: for any finite set of directions $(u,v)$, there exists an arithmetic progression of odd integers $m$ such that \[ 2^{nu}3^{nv}m+1 \] is composite for all $n\ge 1$ and for every chosen direction $(u,v)$;

2. an infinite-prime strengthening: for any such finite ray set, there are infinitely many prime choices of $m$ with the same property;

3. a precise reformulation of the remaining obstruction: the original problem is equivalent to asking whether one can annihilate all primitive rays simultaneously.

This goes well beyond Round 1, which only handled finitely many isolated lattice points.

5) WORK (ROUND-2)

External theorem used (Filaseta).

Fix an integer $A\ge 2$. Then there exists an arithmetic progression of odd positive integers $m$ such that \[ m a^n+1 \] is composite for every integer $a$ with $2\le a\le A$ and every integer $n\ge 1$.

External corollary used (Filaseta + Maynard/Shiu).

For the same fixed $A$, let $K(A)$ denote the set of odd positive integers $m$ such that \[ m a^n+1 \quad \text{and} \quad m+a^n \] are composite for all integers $a$ with $2\le a\le A$ and all $n\ge 1$. Then $K(A)$ contains infinitely many primes; in fact, Filaseta records arbitrarily long strings of consecutive primes in $K(A)$.

Theorem 5.1 (finite-ray annihilation).

Let \[ D\subset \{(u,v)\in \mathbf{Z}_{\ge 0}^2:(u,v)\neq (0,0)\} \] be finite. Then there exists an arithmetic progression of odd positive integers $m$ such that \[ 2^{nu}3^{nv}m+1 \] is composite for every $(u,v)\in D$ and every integer $n\ge 1$.

Proof.

Set \[ A:=\max\{2^u3^v:(u,v)\in D\}. \] Since $D$ is finite, $A$ is a finite positive integer, and because $(u,v)\neq (0,0)$ for every $(u,v)\in D$, we have $2^u3^v\ge 2$, so $A\ge 2$.

Apply Filaseta's theorem to this $A$. Then there exists an arithmetic progression of odd positive integers $m$ such that \[ m a^n+1 \] is composite for every integer $a$ with $2\le a\le A$ and every $n\ge 1$.

Now fix $(u,v)\in D$. The number \[ a:=2^u3^v \] satisfies $2\le a\le A$. Therefore, for every $n\ge 1$, \[ m a^n+1 = m(2^u3^v)^n+1 = m\,2^{nu}3^{nv}+1 \] is composite.

Since this holds for every $(u,v)\in D$, the theorem follows. $\square$

Corollary 5.2 (finite-ray annihilation by infinitely many primes).

With $D$ as above, there are infinitely many prime numbers $m$ such that \[ 2^{nu}3^{nv}m+1 \] is composite for every $(u,v)\in D$ and every $n\ge 1$.

Proof.

Choose $A$ as in Theorem 5.1. By the cited corollary in Filaseta's paper, the set $K(A)$ contains infinitely many primes. Every $m\in K(A)$ satisfies \[ m a^n+1 \text{ composite for all } 2\le a\le A,\ n\ge 1. \] In particular this holds for the finitely many bases $a=2^u3^v$ with $(u,v)\in D$. Hence every prime $m\in K(A)$ has the desired finite-ray annihilation property. $\square$

Proposition 5.3 (primitive-ray decomposition).

Every exponent pair $(k,\ell)\in \mathbf{Z}_{\ge 0}^2\setminus \{(0,0)\}$ can be written uniquely in the form \[ (k,\ell)=n(u,v), \] where $n\ge 1$ and $(u,v)\in \mathbf{Z}_{\ge 0}^2$ is primitive, i.e. \[ \gcd(u,v)=1. \]

Proof.

Let \[ n:=\gcd(k,\ell), \] with the conventions $\gcd(k,0)=k$ and $\gcd(0,\ell)=\ell$. Since $(k,\ell)\neq (0,0)$, we have $n\ge 1$. Define \[ u:=k/n,\qquad v:=\ell/n. \] Then $(u,v)\in \mathbf{Z}_{\ge 0}^2$, $(k,\ell)=n(u,v)$, and $\gcd(u,v)=1$ by construction.

For uniqueness, suppose \[ (k,\ell)=n(u,v)=n'(u',v') \] with both $(u,v)$ and $(u',v')$ primitive. Taking gcds of the coordinates on both sides gives $n=n'$, and then $(u,v)=(u',v')$ follows immediately. $\square$

Sharp reformulation of the remaining obstruction.

By Proposition 5.3, the original problem asks whether there exists an odd $m>1$ such that every value \[ 2^k3^\ell m+1 \] is composite, equivalently: whether one can choose a single $m$ that annihilates every primitive ray \[ \{n(u,v):n\ge 1\}, \qquad \gcd(u,v)=1. \]

Theorem 5.1 and Corollary 5.2 show that every finite collection of primitive rays can be annihilated simultaneously, even by infinitely many prime choices of $m$.

So the unresolved step is now isolated sharply:

Can one pass from finite primitive-ray annihilation to annihilation of the full infinite primitive-ray set?

This is strictly stronger than the Round 1 finite-box perspective, because each ray contains infinitely many exponent pairs.

6) ADVERSARIAL VERIFICATION

I now check the places where the new argument could fail.

(i) Does Theorem 5.1 accidentally miss the pair $(0,0)$?

Yes, it misses it intentionally, because the ray decomposition only applies to nonzero exponent pairs. But for odd $m>1$, \[ 2^0 3^0 m +1 = m+1 \] is even and at least $4$, hence composite. So this is not a real gap.

(ii) Are the axis directions included?

Yes. The primitive rays $(1,0)$ and $(0,1)$ are included in the framework and correspond to the bases $2$ and $3$.

(iii) Is it legitimate to choose $A$ as the maximum of the bases $2^u3^v$?

Yes. Filaseta's theorem works for all integers $a$ with $2\le a\le A$, so it certainly covers the special finitely many integers of the form $2^u3^v$ that I need.

(iv) Does the prime strengthening really follow?

Yes. Filaseta's Corollary 4.4 gives infinitely many primes in $K(A)$, and every such prime belongs to the annihilating set from Theorem 5.1.

(v) Have I solved the original problem?

No. The argument only proves finite-ray annihilation, not annihilation of the full infinite primitive-ray family. That remaining passage is exactly the unresolved step, and I have stated it explicitly.

So the Round 2 work is rigorous and strictly stronger than Round 1, but it still falls short of a full resolution.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict Round 2 advance is the finite-ray annihilation theorem: any finite collection of primitive rays in the $(k,\ell)$-lattice can be annihilated simultaneously, even by infinitely many prime choices of $m$.

This is substantially stronger than the Round 1 finite-box lemma and isolates the exact remaining obstruction: extending from finitely many primitive rays to all primitive rays.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 52%

9) REFERENCES

1. M. Filaseta, Generalized Sierpiński Numbers, arXiv:2305.09219, 2023.
2. T. F. Bloom, Erdős Problem #203, erdosproblems.com, accessed 2026-03-07.