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Math-715-A2 · revision r1 · on Math-715

gpt pro 5.2 attempt on Math-715

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #200: Longest arithmetic progression of primes up to $N$
  2. 1) Formal restatement (quantifiers, definitions, edge cases)
  3. Problem statement.
  4. Edge cases.
  5. 2) Quick literature/context check
  6. 3) Attack plan
  7. 4) Work
  8. Theorem 4.1 (PNT upper bound).
  9. Computational check (moderate $N$).
  10. 5) Verification / consistency checks
  11. 6) Final
  12. UNRESOLVED.
  13. 7) Completion estimate

Erdős Problem #200: Longest arithmetic progression of primes up to $N$

1) Formal restatement (quantifiers, definitions, edge cases)

For $N\ge 2$, let $L(N)$ be the maximum integer $k$ such that there exist integers $a\ge 2$ and $d\ge 1$ with \[ a,\, a+d,\, a+2d,\,\dots,\, a+(k-1)d \ \text{all prime and}\ a+(k-1)d\le N. \] (Equivalently, $L(N)$ is the length of the longest arithmetic progression contained in the set of primes $\le N$.)
Problem statement.
Is it true that \[ L(N)=o(\log N)\qquad (N\to\infty)? \]
Edge cases.
$L(N)$ is nondecreasing in $N$ and $L(N)\ge 1$ always. For $k\ge 3$, any $k$-term progression of odd primes has even common difference $d$.

2) Quick literature/context check

  • The primes contain arbitrarily long arithmetic progressions (Green--Tao theorem) [GreenTao08], so $L(N)\to\infty$, but existing proofs give no usable quantitative growth rate for $L(N)$.
  • A standard application of the prime number theorem (via the Chebyshev function $\vartheta(x)=\sum_{p\le x}\log p\sim x$) yields the upper bound \[ L(N)\le (1+o(1))\log N. \] A complete proof is included below.
  • Heuristics based on treating primes as a random set of density $\approx 1/\log N$ (and refined Hardy--Littlewood type constants) predict that the expected number of $k$-term prime progressions up to $N$ is about \[ \asymp \frac{N^2}{(k-1)(\log N)^k}, \] suggesting that the maximal $k$ satisfies $k\approx 2\log N/\log\log N$ (hence $o(\log N)$). A statement of this heuristic count appears, for instance, in The Prime Pages discussion of prime arithmetic progressions [PrimePagesAP].

3) Attack plan

  1. Prove the known PNT-based upper bound $L(N)\le (1+o(1))\log N$ carefully, since it is explicitly mentioned in the prompt.
  2. For the main open question $L(N)=o(\log N)$, outline why "current technology" (sieve bounds, distribution in residue classes, Green--Tao transference) does not presently deliver such a sharpening.
  3. Provide a quick computational sanity check for moderate $N$.

4) Work

Theorem 4.1 (PNT upper bound).
Let $p_0, p_1,\dots,p_{k-1}$ be a $k$-term arithmetic progression of primes with $p_{k-1}\le N$. Then \[ k\le (1+o(1))\log N\qquad (N\to\infty). \]
Proof: Write the progression as $p_j=a+jd$ with $d\ge 1$.

Step 1: small primes up to $k/2$ divide $d$. Let $q$ be a prime with $q\le k/2$. Suppose for contradiction that $q\nmid d$. Then $d$ is invertible modulo $q$, so the congruence \[ a+jd\equiv 0\pmod q \] has exactly one solution class $j\equiv j_0\pmod q$. Among indices $j\in\\{0,1,\dots,k-1\\}$ there are at least $\lceil k/q\rceil\ge 2$ solutions (since $k\ge 2q$). Hence at least two distinct terms $a+jd$ are divisible by $q$. At most one of them can equal $q$; the other is a multiple of $q$ exceeding $q$, hence composite. This contradicts that all terms are prime. Therefore $q\mid d$ for every prime $q\le k/2$.

Let \[ P(x):=\prod_{q\le x\atop q\text{ prime}} q \] denote the primorial. We have shown $P(k/2)\mid d$, so $d\ge P(k/2)$.

Step 2: almost all primes up to $k$ divide $d$. For large $k$, $P(k/2)$ is enormous, in particular $P(k/2)>k$, so $d>k$. Now let $q$ be a prime with $k/2<q\le k$. If $q\nmid d$, then as above there is at least one index $j\in\\{0,1,\dots,k-1\\}$ such that $a+jd\equiv 0\pmod q$ (since $k\ge q$). That term is a prime divisible by $q$, so it must equal $q$. But $q\le k<d$, so the only term of the progression that could possibly equal $q$ is the first term ($j=0$); all others satisfy $a+jd\ge a+d>d>k\ge q$. Thus, for primes $q$ with $k/2<q\le k$, the only possible exception to "$q\mid d$" is $q=a$ (and this can happen only if $a\le k$). In particular, the set of primes $q\le k$ not dividing $d$ has size at most $1$, so \[ \log d\\ \ge\\ \sum_{q\le k\atop q\text{ prime}}\log q\\ -\\ \log k\\ =\\ \vartheta(k)-\log k, \] where $\vartheta(k):=\sum_{q\le k,\\ q\text{ prime}}\log q$ is Chebyshev's function.

Step 3: apply the prime number theorem. The prime number theorem implies $\vartheta(k)\sim k$ as $k\to\infty$. Hence \[ \log d\ge (1+o(1))k. \] Since $p_{k-1}=a+(k-1)d\le N$, we have $N\ge (k-1)d$, and therefore \[ \log N\ge \log d+\log(k-1)\ge (1+o(1))k. \] Rearranging gives $k\le (1+o(1))\log N$. ∎

Computational check (moderate $N$).
A brute-force search over all prime pairs $(p,q)$ with $p<q\le N$ (testing the progression with step $d=q-p$) gives: \[ L(10^3)=7,\quad L(5\cdot 10^3)=10,\quad L(10^4)=10,\quad L(2\cdot 10^5)=10, \] with a longest example for these ranges being \[ 199,\\ 409,\\ 619,\dots,\\ 2089\qquad (\text{length }10,\\ d=210). \] This is only a sanity check; it is not evidence for asymptotic behaviour.

5) Verification / consistency checks

  • The key modular argument is checked in two regimes: $q\le k/2$ forces $\ge2$ multiples, hence $q\mid d$; $q\in(k/2,k]$ gives at most one multiple, forcing a near-total divisibility of $d$ when $d>k$.
  • The only analytic input is $\vartheta(x)\sim x$, a standard equivalent form of the prime number theorem.

6) Final

UNRESOLVED.
  1. Strongest proved partial result included here. Theorem 4.1 proves the known upper bound $L(N)\le (1+o(1))\log N$ from the prime number theorem. Also $L(N)\to\infty$ by Green--Tao [GreenTao08].
  2. First precise obstacle. To improve the bound to $o(\log N)$ one would need a quantitative upper bound excluding $k$-term prime progressions for $k$ as large as $c\log N$, uniformly in the common difference. Current sieve/analytic tools (and the transference methods in Green--Tao) do not supply such a uniform "union bound" at this scale.
  3. Most plausible next lemma. A plausible route would be a sharp upper bound for the number of $k$-term prime progressions up to $N$ of the form \[ \#\\{(a,d):\\ a,a+d,\dots,a+(k-1)d\le N\text{ all prime}\\}\\ \le\\ N^2\exp(-c k\log\log N) \] for $k$ up to (say) $\varepsilon\log N$, which would imply (by the first moment method) that $L(N)=o(\log N)$. Proving such bounds seems to require substantially stronger uniformity estimates for primes in progressions than are currently available.

7) Completion estimate

\[ \textbf{Completion: }40\%. \] (We can prove the standard PNT upper bound cleanly and explain the heuristic, but the main asymptotic sharpening remains open.)