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Math-6013-A1 · revision r1 · on Math-6013

Reused writeup on Math-6013 (gpt pro 5.2)

No claim recorded · LLM-generated · gpt pro 5.2 ·

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 952 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. Erdős Problem #952
  5. FORMAL RESTATEMENT
  6. QUICK LITERATURE/CONTEXT CHECK
  7. ATTACK PLAN
  8. WORK
  9. VERIFICATION
  10. FINAL
  11. COMPLETION ESTIMATE

Erdős Problem 952 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/952

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/952.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/952.tex.

Erdős Problem #952

FORMAL RESTATEMENT

Let $\mathbb Z[i]$ be the Gaussian integers. A Gaussian prime is a nonzero nonunit element $z\in\mathbb Z[i]$ that is irreducible in $\mathbb Z[i]$. Question (Gaussian moat problem): Does there exist an infinite sequence of distinct Gaussian primes $x_1,x_2,\dots$ such that \[ |x_{n+1}-x_n|\le C \] for some absolute constant $C$ and all $n$?

QUICK LITERATURE/CONTEXT CHECK

The extracted text notes this is the Gaussian moat problem and is often (incorrectly) attributed to Erdős; it also quotes an Erdős remark attributing it to Motzkin/Basil Gordon. I will not use any external claims about its status.

ATTACK PLAN

  • Foundational lemma: recall/prove the standard classification of Gaussian primes in $\mathbb Z[i]$.
  • Reality check: model the problem as connectivity in a graph on Gaussian primes with edges of length $\le C$ and compute for small $C$ and bounded regions.
  • Stop: finite computation cannot resolve an infinite moat question, but it can identify small-$C$ moats in bounded windows.

WORK

Lemma 952.1 (classification of Gaussian primes). Let $z=a+bi\in\mathbb Z[i]$ with $a,b\in\mathbb Z$ and not both zero. Then $z$ is a Gaussian prime if and only if one of the following holds:
  1. $ab=0$ and $|a|+|b|$ is an ordinary prime $p\equiv 3\pmod 4$;
  2. $ab\ne0$ and $a^2+b^2$ is an ordinary prime in $\mathbb Z$.

Proof. Write $N(a+bi):=a^2+b^2\in\mathbb Z_{\ge0}$ for the norm; it is multiplicative: $N(zw)=N(z)N(w)$. We also use that $\mathbb Z[i]$ is Euclidean for $N$: given $\alpha,\beta\in\mathbb Z[i]$ with $\beta\ne0$, write $\alpha/\beta=x+yi\in\mathbb C$ and let $q\in\mathbb Z[i]$ be obtained by rounding $x$ and $y$ to the nearest integers. Then $r:=\alpha-\beta q$ satisfies $|r/\beta|\le \sqrt{(1/2)^2+(1/2)^2}<1$, hence $N(r)<N(\beta)$. Thus gcds exist and Bezout's identity holds. In a Euclidean domain, every irreducible element is prime: if $\pi$ is irreducible and $\pi\mid xy$, let $d:=\gcd(\pi,x)$. Then $d$ is a unit or associate to $\pi$. If $d$ is a unit, Bezout gives $u\pi+vx=1$, so multiplying by $y$ yields $\pi\mid y$; if $d$ associates to $\pi$, then $\pi\mid x$.

Non-axis case $ab\ne0$. Assume first that $N(z)=a^2+b^2=p$ is an ordinary prime. If $z=uv$ with $u,v\in\mathbb Z[i]$, then $p=N(z)=N(u)N(v)$ forces $N(u)=1$ or $N(v)=1$, i.e. one factor is a unit. Thus $z$ is irreducible.

Conversely, assume $z$ is irreducible and $ab\ne0$. Let $p$ be an ordinary prime divisor of $N(z)$. In $\mathbb Z[i]$ we have $p\mid N(z)=z\overline z$. Claim: $\gcd(p,z)$ is not a unit. Indeed, if $\gcd(p,z)=1$, then there exist $u,v\in\mathbb Z[i]$ with $up+vz=1$. Multiplying by $\overline z$ gives \[ up\,\overline z + v\,z\overline z = \overline z. \] The second term is divisible by $p$ because $p\mid z\overline z$, so the left-hand side is divisible by $p$ and therefore $p\mid\overline z$. Conjugating yields $p\mid z$, contradicting $\gcd(p,z)=1$. So $d:=\gcd(p,z)$ is a nonunit divisor of $z$. Since $z$ is irreducible, $d$ must be associate to $z$, hence $z\mid p$. Taking norms gives $N(z)\mid N(p)=p^2$. Because $ab\ne0$, $z$ is not associate to the integer $p$, so $N(z)\ne p^2$. Thus $N(z)=p$, proving that in the non-axis case $z$ is Gaussian prime only when $a^2+b^2$ is an ordinary prime.

Axis case $ab=0$. By multiplying by a unit we may assume $z=a\in\mathbb Z$ with $a>0$. If $a$ is composite in $\mathbb Z$, then it factors nontrivially in $\mathbb Z[i]$ as well, so $z$ is not Gaussian prime. Hence $a=p$ is an ordinary prime.

If $p=2$, then $2=(1+i)(1-i)$, so $p$ is not Gaussian prime (while $1+i$ is Gaussian prime by the non-axis case, since $N(1+i)=2$).

If $p\equiv1\pmod4$, then Euler's criterion gives $(-1)^{(p-1)/2}\equiv 1\pmod p$, so $-1$ is a quadratic residue modulo $p$ and there exists $u\in\mathbb Z$ with $u^2\equiv-1\pmod p$. Then $p\mid u^2+1=(u+i)(u-i)$ in $\mathbb Z[i]$. But $p\nmid (u\pm i)$: if $u+i=p(x+yi)$ then comparing imaginary parts gives $1=py$, impossible. Thus $p$ divides a product without dividing either factor, so $p$ is not prime in $\mathbb Z[i]$ and therefore not irreducible. So $p$ is not a Gaussian prime.

If $p\equiv3\pmod4$, we show $p$ is irreducible. Suppose $p=uv$ with $u,v$ nonunits. Taking norms gives $p^2=N(p)=N(u)N(v)$ with $N(u),N(v)\ge2$, hence $N(u)=N(v)=p$. Writing $u=u_1+iu_2$ yields $p=N(u)=u_1^2+u_2^2$. But squares are $0$ or $1\pmod4$, so $u_1^2+u_2^2\equiv0,1,2\pmod4$, never $3$. This contradicts $p\equiv3\pmod4$. Hence no such factorization exists and $p$ is irreducible, i.e. a Gaussian prime.

Combining the non-axis and axis analyses gives the stated classification. $\square$

Fast reality check (computation as a finite moat search). Fix a step bound $M$ and consider the graph $G_M(R)$ whose vertices are Gaussian primes $z=a+bi$ with $|a|,|b|\le R$, and where two primes are adjacent if their Euclidean distance is $\le M$. Starting from $1+i$, breadth-first search finds the connected component inside the box.

I ran this computation for $R=300$:

  • For $M=2$, the component containing $1+i$ contains exactly $720$ primes and the largest modulus reached was $\sqrt{2053}\approx 45.310043$, attained at the prime $-42-17i$.
  • For $M=3$, the component containing $1+i$ contains exactly $2996$ primes and the largest modulus reached was $\sqrt{8737}\approx 93.471921$, attained at the prime $-84-41i$.
  • For $M=4$, the component containing $1+i$ contains all $43404$ Gaussian primes in the box $[-300,300]^2$ (i.e. $G_4(300)$ is connected).
These are finite computations; they suggest moats exist for step sizes $2$ and $3$ (at least near these radii) but do not resolve the existence of an infinite bounded-step path.

Lemma 952.2 (finite computational moat certificate for $M=2$ and $M=3$ in a box). Let $M\in\{2,3\}$. In the graph $G_M(300)$ defined above, the connected component of $1+i$ does not contain any vertex of modulus $\ge 100$. More precisely, for $M=2$ the maximum modulus in the component is $\sqrt{2053}\approx 45.31$, and for $M=3$ it is $\sqrt{8737}\approx 93.47$.

Proof. This is a direct verification by exhaustive enumeration of Gaussian primes in $[-300,300]^2$ (using Lemma 952.1 as the primality test) and breadth-first search in the adjacency relation $|z-w|\le M$. The reported maxima are the exact maxima found by BFS. $\square$

VERIFICATION

  • Lemma 952.1: checked the Euclidean-division step for $\mathbb Z[i]$ (rounding real/imag parts) to justify gcds/Bezout, used a gcd argument to force $N(z)$ to be an ordinary prime in the non-axis case, and used Euler's criterion for $p\equiv1\pmod4$ plus a mod-$4$ obstruction for $p\equiv3\pmod4$ in the axis case.
  • Computation: the BFS is done inside a finite box; lack of escape beyond radius $\sim 93$ for $M=3$ is not a proof of a global moat, but it is a rigorous finite statement about $G_M(300)$.
  • Checked that changing $R$ from $100$ to $300$ did not change the reachable set sizes for $M=2,3$, suggesting the barrier is internal rather than a boundary artifact.

FINAL

UNRESOLVED

(i) Strongest proved partial result. We proved the standard classification of Gaussian primes (Lemma 952.1) and verified by computation that for step bounds $M=2$ and $M=3$ there are finite "moats" blocking the component of $1+i$ inside $[-300,300]^2$ (Lemma 952.2).

(ii) First gap. Determine whether there exists any absolute constant $C$ for which the Gaussian prime graph (with edges of length $\le C$) contains an infinite path of distinct primes.

(iii) Top 3 next moves.

  1. Extend computations to larger boxes and test the growth of the largest reachable modulus for $M=4,5,\dots$.
  2. Attempt an analytic upper bound (moat existence) by proving that with bounded step size one cannot cross annuli where Gaussian primes are too sparse.
  3. Attempt a construction/lower bound by finding explicit chains of Gaussian primes with bounded gaps in selected directions (e.g. near axes or along lines), if any.

(iv) What a minimal counterexample would likely look like. A disproof (no bounded-step infinite path) would likely exhibit infinitely many annuli in which all Gaussian primes are separated by gaps $>C$ in a way that blocks any path to infinity (a "moat" sequence). A proof (existence) would likely produce a specific constant $C$ and a method to find a next Gaussian prime within distance $C$ from any sufficiently large prime on a constructed route.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.