Erdős Problem #213

Erdős Problem #213 (Round 2)

Question recalled. For each $n\geq 4$, do there exist $n$ points in $\mathbb{R}^2$, no three on a line and no four on a circle, such that all pairwise distances are integers?

1. ROUND-2 OBJECTIVE

Path C: obstruction/structure plus exact verification. Round 1 already had an explicit $4$-point example and had recorded a $7$-point configuration from Kreisel--Kurz, but the $7$-point example was not fully verified and no common structure theorem had been proved. The most promising strict advance is therefore to do two things: prove the standard quadratic-field/characteristic theorem for planar integral point sets, and then turn the recorded $7$-point coordinates into a complete exact proof. This immediately upgrades existence from one hand-checked example at $n=4$ to a rigorously verified existence theorem for every $n\leq 7$.

2. Round-1 FOUNDATION USED

I use the following Round-1 material.

1. Proposition 4.1: an explicit $4$-point general-position integral set.
2. Proposition 4.2: the coordinates of a $7$-point integral set from Kreisel--Kurz were already recorded.
3. The Round-1 gap statement: there was no proof that the $7$-point coordinates really satisfy all $21$ integrality conditions and all general-position checks.

3. NEW INSIGHT / TOOL (ROUND 2)

There are two new pieces of mathematics in this round.

1. A common characteristic theorem: every non-collinear planar integral point set lies, after normalization, in one quadratic field $\mathbb{Q}(\sqrt{k})$, and all its nondegenerate triangles have the same characteristic.
2. A full exact certificate for the Kreisel--Kurz heptagon: all $21$ pairwise distances are listed explicitly, and the no-three-collinear / no-four-concyclic conditions are checked by exact determinant tests in $\mathbb{Q}(\sqrt{2002})$.

4. ATTACK PLAN (ROUND 2)

Round 1 left two exact gaps.

1. It had no structural theorem explaining why the $7$-point coordinates live in $\mathbb{Q}(\sqrt{2002})$.
2. It had no rigorous verification of the full $7$-point example.

So I first prove the characteristic theorem, which sharpens the search space for any future $8$-point construction. I then use exact symbolic computation in the field $\mathbb{Q}(\sqrt{2002})$ to certify all pairwise distances and all general-position conditions for the recorded heptagon.

5. WORK (ROUND 2)

Definition 5.1 (characteristic of an integral triangle). Let $\triangle XYZ$ be a nondegenerate triangle with integer side lengths. By Heron's formula, its area can be written uniquely in the form \[ \operatorname{Area}(\triangle XYZ)=q\sqrt{k}, \] where $q\in \mathbb{Q}_{>0}$ and $k$ is a squarefree positive integer. The integer $k$ is called the characteristic of the triangle.

Theorem 5.2 (common characteristic theorem). Let $S$ be a non-collinear integral point set in the plane. Then there exists a squarefree integer $k>0$ and a Euclidean motion of the plane such that every point of $S$ has coordinates \[ (r_1,r_2\sqrt{k}) \qquad (r_1,r_2\in \mathbb{Q}). \] In particular, every nondegenerate triangle determined by $S$ has characteristic $k$.

Proof. Choose a non-collinear triple $A,B,C\in S$. After translation and rotation we may assume \[ A=(0,0), \qquad B=(d,0), \] where $d=|AB|\in \mathbb{Z}_{>0}$.

Let $P=(x,y)\in S$. Since the distances $|PA|$ and $|PB|$ are integers, say \[ |PA|=a,\qquad |PB|=b, \] we have \[ x^2+y^2=a^2,\qquad (x-d)^2+y^2=b^2. \] Subtracting gives \[ x=\frac{a^2-b^2+d^2}{2d}\in \mathbb{Q}, \] and hence \[ y^2=a^2-x^2\in \mathbb{Q}. \]

Now write \[ C=(u,v),\qquad v\neq 0. \] As above, $u\in \mathbb{Q}$ and $v^2\in \mathbb{Q}_{>0}$, so \[ v=r\sqrt{k} \] for some $r\in \mathbb{Q}_{\neq 0}$ and some squarefree integer $k>0$.

Take any point $P=(x,y)\in S$. Since $|PC|$ is an integer, \[ |PC|^2=(x-u)^2+(y-v)^2 \] is rational. Expanding, \[ (x-u)^2+y^2+v^2-2yv\in \mathbb{Q}. \] The first three terms are rational, so $2yv\in \mathbb{Q}$. Because $v=r\sqrt{k}$ with $r\neq 0$ rational, it follows that \[ y\in \mathbb{Q}\sqrt{k}. \] Thus every point of $S$ has coordinates in $\mathbb{Q}\times \mathbb{Q}\sqrt{k}$.

Finally, the area of any nondegenerate triangle with vertices in $S$ is \[ \frac12\left| \det\begin{pmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{pmatrix}\right|, \] hence a rational multiple of $\sqrt{k}$. Therefore every nondegenerate triangle in $S$ has characteristic $k$. $\square$

Proposition 5.3 (exact verification of the Kreisel--Kurz heptagon). Let $\omega=\sqrt{2002}$ and define \[ \begin{array}{rcl} P_1&=&(0,0),\\[2mm] P_2&=&(22270,0),\\[2mm] P_3&=&\left(\dfrac{26127018}{2227}, \dfrac{932064}{2227}\,\omega\right),\\[3mm] P_4&=&\left(\dfrac{245363}{17}, \dfrac{3144}{17}\,\omega\right),\\[3mm] P_5&=&\left(\dfrac{17615968}{2227}, \dfrac{238464}{2227}\,\omega\right),\\[3mm] P_6&=&\left(\dfrac{56068}{17}, \dfrac{3144}{17}\,\omega\right),\\[3mm] P_7&=&\left(\dfrac{19079044}{2227}, -\dfrac{54168}{2227}\,\omega\right). \end{array} \] Then:

1. all $21$ pairwise distances are integers;
2. no three of the points are collinear;
3. no four of the points are concyclic.

Proof. All calculations are exact in the quadratic field $K=\mathbb{Q}(\omega)$.

Step 1: all pairwise distances are integers. A direct exact simplification of the $21$ squared distances gives the following distance table: \[ \begin{array}{c|cccccc} & P_2 & P_3 & P_4 & P_5 & P_6 & P_7\\ \hline P_1 & 22270 & 22098 & 16637 & 9248 & 8908 & 8636\\ P_2 & & 21488 & 11397 & 15138 & 20698 & 13746\\ P_3 & & & 10795 & 14450 & 13430 & 20066\\ P_4 & & & & 7395 & 11135 & 11049\\ P_5 & & & & & 5780 & 5916\\ P_6 & & & & & & 10744 \end{array} \] Hence every pairwise distance is an integer.

Step 2: no three points are collinear. For distinct $i,j,k$, three points are collinear if and only if \[ \Delta_{ijk}:=\det\begin{pmatrix} x_i & y_i & 1\\ x_j & y_j & 1\\ x_k & y_k & 1 \end{pmatrix}=0. \] Each $\Delta_{ijk}$ lies in $K=\mathbb{Q}(\omega)$. Expanding every determinant in the basis $\{1,\omega\}$, an exact check shows that for all \[ 1\leq i<j<k\leq 7 \] the pair of coefficients is not $(0,0)$. Therefore all $35$ determinants $\Delta_{ijk}$ are nonzero, so no three points are collinear.

Step 3: no four points are concyclic. Four points are concyclic if and only if the determinant \[ \Gamma_{ijkl}:=\det\begin{pmatrix} x_i & y_i & x_i^2+y_i^2 & 1\\ x_j & y_j & x_j^2+y_j^2 & 1\\ x_k & y_k & x_k^2+y_k^2 & 1\\ x_l & y_l & x_l^2+y_l^2 & 1 \end{pmatrix} \] vanishes. Again every $\Gamma_{ijkl}$ lies in $K$, and an exact expansion in the basis $\{1,\omega\}$ shows that all \[ \binom{7}{4}=35 \] such determinants are nonzero. Hence no four points are concyclic.

This proves the proposition. $\square$

Corollary 5.4. For every integer $n$ with $4\leq n\leq 7$, there exists a set of $n$ points in $\mathbb{R}^2$, no three on a line and no four on a circle, such that all pairwise distances are integers.

Proof. For $n=4$, use the Round-1 construction. For $5\leq n\leq 7$, take any $n$-point subset of the heptagon in Proposition 5.3. Pairwise integrality and the two general-position conditions are hereditary under taking subsets. $\square$

6. ADVERSARIAL VERIFICATION

The new work was checked against the main possible defects.

1. Common characteristic theorem. The proof really uses non-collinearity only once, namely to guarantee a point with nonzero $y$-coordinate after normalization. This is valid because the problem concerns general-position sets.
2. Exactness versus numerics. All computations for Proposition 5.3 were exact, not floating-point. Equality in $K=\mathbb{Q}(\omega)$ was tested by comparing the coefficients of $1$ and $\omega$.
3. Subset argument. Distances remain integers on subsets, and any collinear triple or concyclic quadruple in a subset would already violate general position in the full $7$-point set.
4. Interpretation of the field $\mathbb{Q}(\sqrt{2002})$. Theorem 5.2 explains why the coordinates of a planar integral point set naturally live in a common quadratic field, so the appearance of $\sqrt{2002}$ is structural rather than accidental.

7. FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The Round-2 gains over Round 1 are:

1. A full proof of the common characteristic theorem for planar integral point sets.
2. A complete exact verification of the recorded $7$-point Kreisel--Kurz construction.
3. Consequently, a rigorous existence theorem for every $n\leq 7$.

The first unresolved case remains $n=8$.

8. COMPLETION ESTIMATE

COMPLETION: 55%

9. REFERENCES

1. T. Kreisel and S. Kurz, There are integral heptagons, no three points on a line, no four on a circle, Discrete Comput. Geom. 39 $2008$, 786--790.
2. T. F. Bloom, Erdős Problem #213, problem page accessed 2026-03-07.

Erdős Problem #213: independent continuation

Date: 23 September 2026. Source versions read in full: 213.tex in gpt_pro_5.4.

1) OBJECTIVE AND SOURCE AUDIT

The target is an arbitrarily large finite planar set with integral pairwise distances, no three collinear and no four concyclic. I read the supplied characteristic proof and seven-point coordinates. Those are established background, not a new construction. I independently executed all of its finite checks, then derived an exact bounded extension search.

2) EXACT VERIFICATION

Represent each source coordinate as $(x_i,t_i\sqrt{2002})$, with rational $x_i,t_i$. Exact rational arithmetic gives the 21 distances, in lexicographic pair order, \[ \begin{split} &22270,22098,16637,9248,8908,8636,21488,11397,15138,20698,13746,\\ &10795,14450,13430,20066,7395,11135,11049,5780,5916,10744. \end{split} \] All 35 determinants with rows $(x_i,t_i,1)$ are nonzero. All 35 determinants with rows $(x_i,t_i,x_i^2+2002t_i^2,1)$ are nonzero. Multiplying the second column by $\sqrt{2002}$ recovers the geometric determinants, so these exact checks certify both general-position conditions. The companion script verification/root_checks.py contains the rational calculation.

3) WORK: COMPLETE FINITE EXTENSION PARAMETRIZATION

More generally, normalize three old points to $O=(0,0)$, $B=(d,0)$, $C=(u,v\sqrt k)$ with $d>0$, $v\ne0$, and $d,u,v\in\mathbb Q$. Any new point $P=(x,t\sqrt k)$ at integer distances $a,b,c$ from them must satisfy \[ x=\frac{a^2-b^2+d^2}{2d},\qquad t=\frac{a^2+u^2+kv^2-c^2-2ux}{2kv}. \tag{1} \] These follow by subtracting the three squared-distance equations. Conversely, a triple $(a,b,c)$ of positive integers and $1$ gives a valid point at those three distances precisely when $x^2+kt^2=a^2$. Distances to the remaining old points are then tested by checking whether their rational squared distances are integer squares. The two determinant tests above enforce general position after adjoining $P$.

Thus all extensions with $a,b,c\le D$ are covered by at most $D^3$ exact tests. This is completeness for a specified bounded search, not a heuristic coordinate grid. It also avoids assuming the new point has rational Cartesian coordinates when the characteristic is nonsquare.

4) ADVERSARIAL VERIFICATION

The old seven-point determinant checks were run; an exhaustive $D^3$ extension search was not run and no eight-point example is claimed. Arbitrarily large configurations need not extend this particular heptagon, so even proving it unextendible would not decide the problem.

7) FINAL AND REFERENCE

UNRESOLVED. The old certificate is now independently reproducible, and $1$ provides an exact finite search specification. The first unsupplied construction remains eight points. The primary problem record checked via search is https://www.erdosproblems.com/213; it credits the heptagon to Kreisel--Kurz. No literature novelty is claimed.

8) COMPLETION ESTIMATE

COMPLETION: 10% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.

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.