gpt pro 5.4 attempt on Math-5456

Erdős Problem #141: consecutive primes in arithmetic progression

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. Round-1 already isolated the basic primorial divisibility obstruction on the common difference. The natural next question is whether local congruence considerations could settle the problem. The most promising strict advance is therefore to determine exactly how far those congruence obstructions can go.

2) Round-1 foundation used

I use the following Round-1 results.

1. The Round-1 primorial divisibility lemma: if $a,a+d,\dots,a+(k-1)d$ are primes and all exceed $k$, then every prime $q\le k$ divides $d$.
2. The Round-1 verification of explicit CPAP examples for $k=3,4,5,6$.
3. The Round-1 literature summary that existence is known computationally for $k\le 10$, but remains open in general and open even for infinitude when $k=3$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a global local-admissibility theorem. After imposing the Round-1 necessary divisibility condition $k\#\mid d$, the prime positions \[ a,\ a+d,\ \dots,\ a+(k-1)d \] still pass every local congruence test simultaneously. In Hardy--Littlewood language, the primorial-step pattern itself is admissible.

This rules out an entire class of disproof strategies: one cannot prove nonexistence of CPAP-$k$ merely by checking divisibility of the designated prime positions modulo primes.

4) ATTACK PLAN (ROUND-2)

After Round-1, the remaining gap was this: the primorial lemma gives a necessary congruence condition on $d$, but does not say whether that condition is already "close to impossible" or in fact perfectly compatible with all local prime obstructions.

I therefore prove the following sharper statement.

For the step size $D_k:=\prod_{p\le k} p$, the $k$ linear forms \[ L_i(n)=n+iD_k\qquad (0\le i\le k-1) \] are admissible at every prime.

That shows the designated prime positions themselves encounter no local obstruction whatsoever.

5) WORK (ROUND-2)

Claim 141.1 (the primorial-step pattern is admissible). Let \[ D_k:=\prod_{p\le k} p. \] For each $i=0,1,\dots,k-1$, define the linear form \[ L_i(n)=n+iD_k. \] Then for every prime $q$, the set of residue classes \[ \{-iD_k \pmod q: 0\le i\le k-1\} \] is a proper subset of $\mathbf Z/q\mathbf Z$. Equivalently, for every prime $q$ there exists a residue class $n\pmod q$ such that \[ q \nmid L_i(n) \qquad \text{for all } i=0,1,\dots,k-1. \] Thus the $k$-tuple $\{L_0,\dots,L_{k-1}\}$ is admissible in the Hardy--Littlewood sense.

Proof. Fix a prime $q$.

Case 1: $q\le k$. Then $q\mid D_k$, so for every $i$ we have \[ L_i(n)=n+iD_k\equiv n \pmod q. \] Hence all $k$ forms are simultaneously divisible by $q$ only when $n\equiv 0\pmod q$. So there is exactly one forbidden residue class modulo $q$, and therefore the forbidden set is a proper subset of $\mathbf Z/q\mathbf Z$.

Case 2: $q>k$. Now $q\nmid D_k$, so $D_k$ is invertible modulo $q$. Therefore the residues \[ 0,\ D_k,\ 2D_k,\ \dots,\ (k-1)D_k \pmod q \] are pairwise distinct. Equivalently, the forbidden residues \[ -iD_k \pmod q \qquad (0\le i\le k-1) \] are pairwise distinct. There are exactly $k$ of them. Since $q>k$, these $k$ residues cannot exhaust all $q$ residue classes modulo $q$. So again the forbidden set is proper.

In both cases there exists at least one residue class $n\pmod q$ for which no $L_i(n)$ is divisible by $q$. Hence the tuple is admissible at every prime $q$. $\square$

Corollary 141.2. No disproof of the existence of CPAP-$k$ can be obtained solely from local congruence obstructions on the $k$ designated prime positions.

Proof. Round-1 already showed that any genuine prime progression of length $k$ with all terms $>k$ must have common difference divisible by $D_k$. Claim 141.1 shows that once this mandatory divisibility is imposed, the $k$ designated prime positions still satisfy every local admissibility condition at every prime. Therefore local divisibility of the designated prime positions cannot force nonexistence. $\square$

Corollary 141.3. For every finite set $S$ of primes, there are infinitely many integers $a$ such that \[ \gcd\left(\prod_{i=0}^{k-1}(a+iD_k),\ \prod_{q\in S} q\right)=1. \]

Proof. For each $q\in S$, choose a good residue class modulo $q$ given by Claim 141.1. The Chinese remainder theorem then yields a residue class modulo $\prod_{q\in S} q$ satisfying all these conditions simultaneously. That residue class contains infinitely many integers $a$. $\square$

6) ADVERSARIAL VERIFICATION

1. Does this solve the CPAP problem? No. The theorem controls only the designated prime positions. A CPAP-$k$ also requires that every integer between successive terms be composite, and nothing here addresses that global/consecutive condition.

2. Did I use the Round-1 primorial lemma correctly? Yes. The new theorem is explicitly designed to start from that lemma's necessary condition $D_k\mid d$ and then show that there is no further local obstruction at the prime positions.

3. Could the admissibility statement fail for $q=k$? No. If $q\le k$, then because $q\mid D_k$ all forms are congruent to $n$ modulo $q$, so only the single residue class $n\equiv 0\pmod q$ is forbidden.

4. Audit of Round-1 computations. I rechecked the explicit Round-1 CPAP examples for $k=3,4,5,6$ by direct primality testing and by enumerating all primes in the interval from the first to the last term. No flaw was found there.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

Round-2 gives a genuine new structural theorem: after imposing the mandatory primorial divisibility from Round-1, the prime positions of a length-$k$ arithmetic progression remain admissible at every prime. So purely local congruence arguments on the designated prime positions cannot settle CPAP-$k$.

The existence question for CPAP-$k$ itself remains open.

8) COMPLETION ESTIMATE

COMPLETION: 34%

9) REFERENCES

• Erdős Problems database, Problem #141 (status page checked March 7, 2026).
• Chris Caldwell, Consecutive Primes in Arithmetic Progression, Prime Pages.
• OEIS A006560 (smallest starting prime for $n$ consecutive primes in arithmetic progression).