gpt pro 5.2 attempt on Math-719

Erdős Problem #213

1. Formal Restatement

For an integer $n\ge 4$, ask whether there exists a set of points \[S=\{p_1,\dots,p_n\}\subset\mathbb{R}^2\] such that:

1. Integral distances: for all $1\le i<j\le n$, $d(p_i,p_j)\in\mathbb{Z}_{>0}$.
2. No three collinear: for all distinct $i,j,k$, the points $p_i,p_j,p_k$ do not lie on a single line.
3. No four concyclic: for all distinct $i,j,k,\ell$, the points $p_i,p_j,p_k,p_\ell$ do not lie on a single circle.

The natural strengthened interpretation (and the one that makes this a genuine "Erdős problem") is:

Do such configurations exist for arbitrarily large $n$ (equivalently, for every $n$ beyond some point)?

Indeed, if a configuration exists for some $N$, then any subset of size $n\le N$ is again in general position and still has all distances integral.

2. Quick Literature/Context Check

The prompt states:

• Anning--Erdős proved there is no infinite set in $\mathbb{R}^2$ with all pairwise distances integral.
• There are explicit finite constructions: $n=5$ (Harborth) and $n=7$ (Kreisel--Kurz).
• Stronger upper bounds are known conditionally (e.g. Bombieri--Lang), and strong "sparseness" statements are known unconditionally for sets inside $[-N,N]^2$ under the same general-position constraints.

This indicates that the existence of such sets for $n\ge 8$ is open.

3. Attack Plan

Proof approach (existence for all large $n$). Try to construct an infinite family of integral distance sets in general position. Typical geometric strategies would be:

• Start from one integral set and add points via circle-circle intersections, ensuring all new distances are integers.
• Use parameterizations of integral triangles sharing a common "characteristic" (as suggested by known constructions), so that many distances become integral in a coordinated way.

Disproof approach (nonexistence beyond some $n_0$). Try to prove an unconditional absolute upper bound on $|S|$ under the general-position assumptions. Possible routes:

• Incidence geometry / additive combinatorics controlling repeated distances.
• Diophantine geometry: show that too many integer distance constraints force collinearity or cocircularity.

4. Work

I cannot resolve the existence question for unbounded $n$, but I can:

• Give a fully explicit construction for $n=4$ satisfying all constraints (verified by hand).
• Record an explicit known construction for $n=7$ (hence also for all $n\le 7$) using published coordinates.

Proposition 4.1 (Explicit $n=4$ example in general position with integral distances). Let \[p_1=(0,0),\quad p_2=(0,16),\quad p_3=(6,8),\quad p_4=(15,8).\] Then:

1. all six pairwise distances are integers,
2. no three of the points are collinear,
3. the four points are not concyclic.

Proof. Step 1: integer distances. Compute squared distances:

• $d(p_1,p_2)^2=(0-0)^2+(0-16)^2=256$, so $d(p_1,p_2)=16$.
• $d(p_3,p_4)^2=(6-15)^2+(8-8)^2=81$, so $d(p_3,p_4)=9$.
• $d(p_1,p_3)^2=(0-6)^2+(0-8)^2=36+64=100$, so $d(p_1,p_3)=10$.
• $d(p_2,p_3)^2=(0-6)^2+(16-8)^2=36+64=100$, so $d(p_2,p_3)=10$.
• $d(p_1,p_4)^2=(0-15)^2+(0-8)^2=225+64=289$, so $d(p_1,p_4)=17$.
• $d(p_2,p_4)^2=(0-15)^2+(16-8)^2=225+64=289$, so $d(p_2,p_4)=17$.

Hence all distances are integers.

Step 2: no three collinear. We check each potential collinear triple by slopes (or signed area). For example:

• $p_1,p_2$ lie on the vertical line $x=0$, but $p_3$ and $p_4$ have $x\neq 0$.
• $p_3,p_4$ lie on the horizontal line $y=8$, but $p_1,p_2$ have $y\neq 8$.
• The line through $p_1=(0,0)$ and $p_3=(6,8)$ has slope $8/6=4/3$; the point $p_4=(15,8)$ does not satisfy $8=(4/3)\cdot 15$.
• The line through $p_2=(0,16)$ and $p_3=(6,8)$ has slope $(8-16)/(6-0)=-8/6=-4/3$; the point $p_4=(15,8)$ does not satisfy $8-16=(-4/3)\cdot 15$.

Thus no three are collinear.

Step 3: not concyclic. A circle through three noncollinear points is unique. Consider the circle through $p_1,p_3,p_4$. The segment $p_3p_4$ is horizontal with midpoint $m_{34}=(10.5,8)$, so its perpendicular bisector is the vertical line $x=10.5=21/2$. The segment $p_1p_3$ has midpoint $m_{13}=(3,4)$ and slope $4/3$, so the perpendicular bisector has slope $-3/4$ and equation \[y-4=-\frac{3}{4}(x-3). \] Intersecting with $x=21/2$ gives \[y-4=-\frac{3}{4}\left(\frac{21}{2}-3\right)=-\frac{3}{4}\cdot\frac{15}{2}=-\frac{45}{8},\quad\text{so }y=-\frac{13}{8}. \] Hence the circumcenter of $\triangle(p_1p_3p_4)$ is \[O=\left(\frac{21}{2},-\frac{13}{8}\right). \] Its radius squared is \[R^2=d(O,p_1)^2=\left(\frac{21}{2}\right)^2+\left(-\frac{13}{8}\right)^2=\frac{441}{4}+\frac{169}{64}=\frac{7225}{64}. \] Now compute the squared distance from $O$ to $p_2=(0,16)$: \[d(O,p_2)^2=\left(\frac{21}{2}\right)^2+\left(16+\frac{13}{8}\right)^2=\frac{441}{4}+\left(\frac{141}{8}\right)^2=\frac{7056}{64}+\frac{19881}{64}=\frac{26937}{64} \neq \frac{7225}{64}=R^2. \] So $p_2$ is not on the circle through $p_1,p_3,p_4$, hence the four points are not concyclic. $\square$

Proposition 4.2 (Explicit $n=7$ example, hence existence for all $n\le 7$). Let $\omega:=\sqrt{2002}$ and consider the seven points \[ \begin{array}{rcl} P_1&=&(0,0),\\ P_2&=&(22270,0),\\ P_3&=&\Bigl(\frac{26127018}{2227},\ \frac{932064}{2227}\,\omega\Bigr),\\ P_4&=&\Bigl(\frac{245363}{17},\ \frac{3144}{17}\,\omega\Bigr),\\ P_5&=&\Bigl(\frac{17615968}{2227},\ \frac{238464}{2227}\,\omega\Bigr),\\ P_6&=&\Bigl(\frac{56068}{17},\ \frac{3144}{17}\,\omega\Bigr),\\ P_7&=&\Bigl(\frac{19079044}{2227},\ -\frac{54168}{2227}\,\omega\Bigr). \end{array} \] These are real points (coordinates in $\mathbb{Q}(\omega)$). A direct computation (as given in the source of this construction) yields that all $\frac{7\cdot 6}{2}=21$ pairwise distances are integers, and the set is in general position (no three collinear, no four concyclic). Consequently, for every $n\le 7$ there exists such a configuration (take any $n$-point subset).

One sample distance check. For instance, between $P_1$ and $P_2$, \[d(P_1,P_2)=22270\in\mathbb{Z}. \] Between $P_4$ and $P_6$ we have the same $y$-coordinate and \[\frac{56068}{17}-\frac{245363}{17}=\frac{-189295}{17}=-11135, \] so $d(P_4,P_6)=11135\in\mathbb{Z}$. $\square$

Sticking point for #213. The known examples show existence up to $n=7$, but there is no general mechanism here to extend a configuration of size $n$ to one of size $n+1$ while preserving both integrality of all new distances and the general-position constraints.

5. Verification

• Proposition 4.1 was checked entirely by explicit arithmetic.
• Proposition 4.2: the coordinates are real and the sample computations illustrate how integrality can be checked; a full verification requires finitely many additional distance computations plus checking all $35$ collinearity tests and all $35$ concyclicity tests.

6. FINAL

UNRESOLVED.

1. (i) Strongest fully proved partial result obtained here:
   • An explicit configuration of $4$ points in $\mathbb{R}^2$ in general position with all pairwise distances integral (Proposition 4.1).
   • An explicit $7$-point configuration is recorded (Proposition 4.2), implying existence for all $n\le 7$.
2. (ii) First gap blocking completion: No argument is provided that constructs such configurations for arbitrarily large $n$ or rules them out beyond some $n_0$. In particular, the case $n=8$ is not resolved here.
3. (iii) Most promising next moves:
   • Attempt to identify an algebraic/parametric "gluing" operation that adds a new point at integer distances to an existing integral set while keeping general position.
   • Pursue unconditional upper bounds using incidence geometry and rigidity of integer distance constraints (aiming to force collinearity/cocircularity when $|S|$ is large).
   • Use modern structural theorems on integer distance sets (sparseness/structure dichotomies) to exclude large general-position configurations.
4. (iv) Conjectured final answer and minimal counterexample shape: I conjecture the answer is NO for all sufficiently large $n$ (i.e. there is a maximal size). A minimal counterexample to this conjecture would be a general-position integral set of size $8$; such a set would likely have very large diameter and live in a quadratic field $\mathbb{Q}(\sqrt{k})$ with a common "characteristic" shared by all triangles, as suggested by known constructions.

7. Completion Estimate

35%. Small-$n$ existence is demonstrated explicitly (including $n=4$ by hand and a recorded $n=7$ example), but the main open direction---existence or nonexistence for unbounded $n$---remains untouched at a structural level.