Erdős Problem #141

Erdős Problem #141: consecutive primes in arithmetic progression

1) ROUND-2 OBJECTIVE

2) Round-1 foundation used

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)

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)

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.

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

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

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).

Erdős Problem #141: independent continuation

1) TARGET AND SOURCE AUDIT

All supplied versions considered: 141.tex. The source verifies admissibility of the designated prime positions with primorial step. I add the missing local treatment of every undesired intervening position; the final primality step remains explicitly conditional.

2) WORK

The desired linear forms $Qt+a+iD$, $0\le i<k$, are admissible. If a prime $p\mid D$, every form is $1\pmod p$. If $p=q_r$, the $i$th form is $(iD-r)\pmod p$, which is nonzero because $0<|iD-r|\le L<p$. For all other primes $p$, $Q$ is invertible modulo $p$, and $p>k$; each form excludes one residue of $t$, so at most $k<p$ residues are excluded.

Consequently the prime-tuples conjecture for these $k$ admissible linear forms would imply infinitely many consecutive-prime progressions of length $k$, even with common difference $D$. This is a conditional implication, not an application of a proved prime-tuples theorem.

3) ADVERSARIAL VERIFICATION

4) FINAL

5) SOURCES CHECKED

6) COMPLETION ESTIMATE

Problem 141: consecutive primes in arithmetic progression

1) FORMAL RESTATEMENT

Let $p_1=2<p_2=3<p_3=5<\cdots$ be the increasing sequence of primes. Fix an integer $k\ge 3$.

The question is: for each fixed $k\ge 3$, does there exist a CPAP-$k$? A stronger open question (mentioned in the prompt) is whether there are infinitely many CPAP-$k$ for fixed $k$ (open even for $k=3$).

2) QUICK LITERATURE / CONTEXT CHECK

• Green--Tao proved that the primes contain arbitrarily long arithmetic progressions, but their progressions need not be consecutive primes.(See [GreenTao2008] and background summaries such as [ErdosProblems141].)

• The "consecutive" requirement is much stronger and is open in general. Computational searches have exhibited examples for small $k$ (the prompt mentions verification for $k\le 10$; this is often recorded under the name CPAP-$k$).(See [WikiCPAP,T5KCPAP].)

• Even for $k=3$, infinitude is open: $p_n,p_{n+1},p_{n+2}$ are in arithmetic progression iff consecutive prime gaps satisfy $p_{n+1}-p_n = p_{n+2}-p_{n+1}$.

3) ATTACK PLAN

There are two natural directions.

1. Obstruction search. Prove necessary congruence/divisibility constraints on the common difference $d$ of any prime AP of length $k$. This can show such progressions must be extremely sparse, but does not by itself rule them out.

2. Existence/infinitude via prime patterns. Heuristically, Hardy--Littlewood type conjectures for prime $k$--tuples suggest infinitely many CPAP-$k$, but proving even CPAP-$3$ infinitude would require breakthroughs on prime gaps/patterns far beyond current technology.

In the "work" below I (i) prove a standard "primorial divides $d$" lemma, and (ii) give explicit small-$k$ examples (with a computational verification of primality and consecutiveness).

4) WORK

4.1 A necessary divisibility condition on the common difference

4.2 Explicit examples for small $k$

The following explicit CPAP-$k$ examples are recorded in standard tables.(See [WikiCPAP,T5KCPAP].) I also verified (by trial division primality testing and checking for intervening primes) that each listed tuple consists of consecutive primes.

\begin{align*} \text{CPAP-3:}\quad & 3,5,7 && (d=2).\\ \text{CPAP-4:}\quad & 251,257,263,269 && (d=6).\\ \text{CPAP-5:}\quad & 9843019,9843049,9843079,9843109,9843139 && (d=30).\\ \text{CPAP-6:}\quad & 121174811,121174841,121174871,121174901,121174931,121174961 && (d=30). \end{align*}

5) VERIFICATION / EDGE CASES

• Lemma Lemma (Primorial divisibility unless a small prime appears) has a small-prime exception: if $q\le k$ is itself one of the terms, then $q\nmid d$ is possible (e.g. progressions starting at $2$ can have odd $d$). For CPAP-$k$ with large $k$ this exception is irrelevant because the terms are enormous.

• The examples in §4.2 are genuinely "consecutive" primes: between successive terms $a+id$ and $a+(i+1)d$ there is no other prime.

6) FINAL

UNRESOLVED

1. Farthest point reached. I proved the primorial divisibility obstruction (Lemma Lemma (Primorial divisibility unless a small prime appears)) and verified concrete CPAP-$k$ examples for $k\le 6$.

2. Best partial lemma. Lemma Lemma (Primorial divisibility unless a small prime appears): for any prime AP of length $k$, every prime $q\le k$ divides the common difference unless $q$ appears as a term.

3. Smallest missing step. To prove existence (or infinitude) of CPAP-$k$ would require controlling not just primes in patterns but also exclusion of primes in the intervals between successive terms (consecutive prime gaps), a problem currently out of reach even for $k=3$.

4. Completion estimate. A proof seems to require breakthroughs comparable to (or stronger than) major open problems on prime $k$--tuples and structured prime gaps; no known method approaches CPAP-$k$ infinitude.