Erdős Problem #212

Erdős Problem #212 (Round 2)

Question recalled. Does there exist a dense set $S\subset \mathbb{R}^2$ such that every pairwise distance between points of $S$ is rational?

1. ROUND-2 OBJECTIVE

Path C: obstruction/structure. Round 1 already identified the core extension gap, so the most promising strict advance is not to restart the existence search but to prove unconditional theorems ruling out two major classes of candidate counterexamples. In this round I prove that an infinite rational distance set cannot have infinitely many points on a line or on a circle unless all but at most four points lie there. I also prove a quadratic-field normal form for every non-collinear rational distance set.

2. Round-1 FOUNDATION USED

I use the following Round-1 results without reproving them.

1. Lemma 4.1: every rational distance set in $\mathbb{R}^2$ is countable.
2. Propositions 4.2 and 4.3: there are dense rational-distance sets on a line and on a circle.
3. The Round-1 gap statement: no mechanism was available to add a point near an arbitrary target while preserving rationality of all distances.

3. NEW INSIGHT / TOOL (ROUND 2)

There are three genuinely new ingredients in this round.

1. A quadratic-field normal form: after a rational similarity, every non-collinear rational distance set lies in $\mathbb{Q}\times \mathbb{Q}\sqrt{k}$ for one fixed squarefree integer $k$.
2. A line obstruction: if infinitely many points of a rational distance set lie on one line, then at most four points lie off that line.
3. A circle obstruction: by rational-distance-preserving inversion, the same bound holds for a circle.

The last two statements imply a new corollary directly relevant to the dense-set problem: a dense rational distance set, if it exists at all, must meet every line and every circle in only finitely many points.

4. ATTACK PLAN (ROUND 2)

Round 1 had only a weak structural fact (countability). The exact missing ingredient was a theorem showing that natural "one-dimensional skeleton" constructions cannot be dense in the plane. I therefore prove, in order:

1. a coordinate rigidity lemma for rational distance sets;
2. a sharp line theorem using a hyperelliptic-curve argument and Faltings' theorem;
3. a circle theorem by inversion around a point of the set.

This overcomes a concrete Round-1 obstacle: any construction built from infinitely many points on one line or one circle plus finitely many auxiliary points is now rigorously ruled out.

5. WORK (ROUND 2)

Lemma 5.1 (quadratic-field normal form). Let $S$ be a rational distance set containing three non-collinear points. Then there exist a squarefree integer $k>0$ and a similarity of the plane whose scale factor is rational such that the image of $S$ is contained in \[ \{(r_1,r_2\sqrt{k}) : r_1,r_2\in \mathbb{Q}\}. \]

Proof. Choose distinct points $A,B\in S$. Since $d(A,B)\in \mathbb{Q}_{>0}$, after a translation, a rotation, and then a scaling by the rational factor $1/d(A,B)$, we may assume \[ A=(0,0), \qquad B=(1,0). \] Let $P=(x,y)$ be any point of the transformed set. Write \[ a=d(P,A)\in \mathbb{Q}, \qquad b=d(P,B)\in \mathbb{Q}. \] Then \[ x^2+y^2=a^2, \qquad (x-1)^2+y^2=b^2. \] Subtracting gives \[ x=\frac{a^2-b^2+1}{2}\in \mathbb{Q}, \] and therefore \[ y^2=a^2-x^2\in \mathbb{Q}. \]

Now choose a point $C=(u,v)$ of the set with $v\neq 0$; this is possible because the set contains a non-collinear triple. Since $u\in \mathbb{Q}$ and $v^2\in \mathbb{Q}_{>0}$, we may write \[ v=r\sqrt{k} \] with $r\in \mathbb{Q}_{\neq 0}$ and $k>0$ squarefree.

Take any point $P=(x,y)$ in the set. Because $d(P,C)\in \mathbb{Q}$, \[ d(P,C)^2=(x-u)^2+(y-v)^2 \] is rational. Expanding, \[ (x-u)^2+y^2+v^2-2yv \in \mathbb{Q}. \] The first three terms are rational, so $2yv\in \mathbb{Q}$. Since $v=r\sqrt{k}$ with $r\in\mathbb{Q}_{\neq 0}$, it follows that \[ \frac{y}{\sqrt{k}}\in \mathbb{Q}. \] Hence $y\in \mathbb{Q}\sqrt{k}$, as required. $\square$

Lemma 5.2 (inversion preserves rationality). Let $S$ be a rational distance set, let $O\in S$, and let $\rho\in \mathbb{Q}_{>0}$. Define the inversion \[ I(P)=O+\frac{\rho^2(P-O)}{|P-O|^2} \] for $P\neq O$. Then $I(S\setminus\{O\})$ is again a rational distance set.

Proof. For distinct $P,Q\neq O$ the standard inversion formula gives \[ |I(P)-I(Q)|=\frac{\rho^2\,|P-Q|}{|P-O|\,|Q-O|}. \] Here $|P-Q|$, $|P-O|$, and $|Q-O|$ are all rational because $O,P,Q\in S$. Since $\rho^2\in \mathbb{Q}$, the right-hand side is rational. $\square$

Theorem 5.3 (line theorem). Let $S$ be a rational distance set in the plane. If $S$ has infinitely many points on a line $L$, then at most four points of $S$ lie off $L$.

Proof. Assume for contradiction that $S\cap L$ is infinite and that there are at least five points of $S$ off $L$.

Choose distinct points $A,B\in S\cap L$. After a translation, rotation, and rational scaling as in Lemma 5.1, we may assume \[ L=\{(x,0):x\in \mathbb{R}\},\qquad A=(0,0),\qquad B=(1,0). \] If $(x,0)\in S\cap L$, then, exactly as in Lemma 5.1, \[ x=\frac{d((x,0),A)^2-d((x,0),B)^2+1}{2}\in \mathbb{Q}. \] So the set \[ X:=\{x\in \mathbb{Q} : (x,0)\in S\cap L\} \] is infinite.

Every off-line point has the form $P=(a,b)$ with $b\neq 0$, where $a\in \mathbb{Q}$ and $b^2\in \mathbb{Q}$ by the same calculation as in Lemma 5.1. Points reflected across $L$ give the same pair $(a,b^2)$, so a reflection class contains at most two points. Since there are at least five off-line points, there are at least three distinct reflection classes. Choose representatives \[ P_j=(a_j,b_j),\qquad j=1,2,3, \] from three distinct reflection classes. Then the quadratic polynomials \[ f_j(x):=(x-a_j)^2+b_j^2 \] lie in $\mathbb{Q}[x]$, have degree $2$, and are pairwise distinct. Because $b_j\neq 0$, each $f_j$ has two distinct non-real roots $a_j\pm i b_j$, so each $f_j$ is squarefree. Distinct reflection classes give distinct pairs $(a_j,b_j^2)$, hence distinct root sets, so the product \[ F(x):=f_1(x)f_2(x)f_3(x) \] is squarefree of degree $6$.

For every $x\in X$, each $f_j(x)$ is the square of the rational distance from $(x,0)$ to $P_j$, hence each $f_j(x)$ is a rational square. Therefore \[ y^2=F(x) \] has a rational solution $y$ for every $x\in X$. So the affine curve \[ C:\ y^2=F(x) \] has infinitely many rational points.

Now $F$ is squarefree of degree $6$, so the smooth projective model of $C$ is a hyperelliptic curve of genus \[ g=\frac{6-2}{2}=2. \] By Faltings' theorem, a curve over $\mathbb{Q}$ of genus at least $2$ has only finitely many rational points. This contradicts the infinitude of $X$.

Therefore at most four points of $S$ lie off $L$. $\square$

Corollary 5.4. If a dense rational distance set $S\subset \mathbb{R}^2$ exists, then $S$ meets every line in only finitely many points.

Proof. If some line contained infinitely many points of $S$, then by Theorem 5.3 all but at most four points of $S$ would lie on that line. The closure of such a set is contained in that line together with finitely many isolated points, so it cannot be all of $\mathbb{R}^2$. $\square$

Theorem 5.5 (circle theorem). Let $S$ be a rational distance set in the plane. If $S$ has infinitely many points on a circle $C$, then at most four points of $S$ lie off $C$.

Proof. Choose a point $O\in S\cap C$; this is possible because $S\cap C$ is infinite. Apply inversion centered at $O$ with radius $\rho=1$. By Lemma 5.2, \[ S':=I(S\setminus\{O\}) \] is a rational distance set.

A circle through the inversion center maps to a line, so $C\setminus\{O\}$ maps to a line $L$. Since $S\cap C$ is infinite, the set $S'\cap L$ is infinite. By Theorem 5.3, at most four points of $S'$ lie off $L$. These correspond exactly to the points of $S$ off $C$. Hence at most four points of $S$ lie off $C$. $\square$

Corollary 5.6. If a dense rational distance set $S\subset \mathbb{R}^2$ exists, then $S$ meets every circle in only finitely many points.

Proof. Exactly as in Corollary 5.4, using Theorem 5.5 instead of Theorem 5.3. $\square$

6. ADVERSARIAL VERIFICATION

I checked the new argument against the main failure modes.

1. Repeated quadratic factors. This is the subtle point in the line theorem. If two off-line points are reflections across $L$, they produce the same quadratic $(x-a)^2+b^2$. I handled this by passing to reflection classes. Since each class has size at most $2$, five off-line points force at least three distinct classes.
2. Rationality of coefficients. After the normalization $A=(0,0)$ and $B=(1,0)$, every point $(a,b)$ of the rational distance set satisfies $a\in \mathbb{Q}$ and $b^2\in \mathbb{Q}$, so the curve really is defined over $\mathbb{Q}$.
3. Genus computation. The product polynomial has degree $6$ and is squarefree, so the hyperelliptic genus formula gives genus $2$ exactly.
4. Inversion. The inversion center $O$ is removed, but that does not affect the exceptional-point count because $O$ lies on the original circle.
5. Relevance to density. The corollaries do not solve the Erdős--Ulam problem, but they do sharply rule out any construction that uses infinitely many points on one line or one circle.

7. FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The new Round-2 advances over Round 1 are:

1. Every non-collinear rational distance set admits a common quadratic-field normal form.
2. If infinitely many points lie on one line, then all but at most four points lie on that line.
3. If infinitely many points lie on one circle, then all but at most four points lie on that circle.
4. Therefore any dense rational distance set, if it exists, must intersect every line and every circle only finitely many times.

The first unresolved gap is now very sharp: one would have to rule out (or construct) a countable rational distance set which is dense in $\mathbb{R}^2$ and has only finite intersection with every line and every circle.

8. COMPLETION ESTIMATE

COMPLETION: 45%

9. REFERENCES

1. G. Faltings, Endlichkeitssätze für abelsche Varietäten über Zahlkörper, Invent. Math. 73 $1983$, 349--366.
2. T. F. Bloom, Erdős Problem #212, problem page accessed 2026-03-07.

OBJECTIVE AND SOURCE REVIEW

Erdős #212; independent attempt, 2026-09-23. Can a rational-distance set be dense in the plane? I read all of 212.tex: its quadratic-field normalization, genus-two line obstruction and inversion argument. Here is an elementary obstruction to constructing such a set as an additive group, requiring no algebraic-geometry theorem.

WORK: ADDITIVE CLOSURE IS IMPOSSIBLE

An additive subgroup $G\subset\mathbb R^2$ whose elements all have rational norm lies on a line. Suppose otherwise that $u,v\in G$ are independent. Write $a=|u|\in\mathbb Q_{>0}$, $b=u\cdot v\in\mathbb Q$, $c=|v|^2\in\mathbb Q$. Rationality of $b$ follows by subtracting $|u+v|^2-|u|^2-|v|^2$. Choose a positive integer $K$ such that $A=Ka$, $B=Kb/a$, and $K^2c$ are integers. Then, for every integer $m\ge1$, \[ (K|mu+v|)^2=(Am+B)^2+D,\qquad D=K^2c-B^2>0. \] Here $D$ is a positive integer, by independence and Cauchy--Schwarz. The left side is a rational square and an integer, so it must be the square of an integer (reduce a rational square root to lowest terms). For sufficiently large $m$, however, \[ (Am+B)^2<(Am+B)^2+D<(Am+B+1)^2, \] an impossibility. This proves the claim.

ADVERSARIAL VERIFICATION AND REMAINING GAP

Only closure under $mu+v$ is used; general rational-distance sets have no such closure. The integer scaling is chosen once, independent of $m$. The positivity $D>0$ fails exactly in the collinear case, where rational additive groups can indeed be dense on a line. The lemma rules out dense additive groups and their translates as constructions, but gives no obstruction to irregular countable sets lacking additive closure. No extension theorem or unrestricted nonexistence result is obtained.

SOURCES CHECKED

Complete local source: gpt_pro_5.4/212.tex. Official indexed statement checked at https://www.erdosproblems.com/212, 2026-09-23. The subgroup proof above is self-contained and is not claimed to be new in the literature.

FINAL

UNRESOLVED. Additive subgroup constructions are excluded; arbitrary dense sets remain untreated.

COMPLETION ESTIMATE

Conservative subjective progress toward planar density. COMPLETION: 10%

Erdős Problem #212

1. Formal Restatement

Let $d(\cdot,\cdot)$ denote the Euclidean distance on $\mathbb{R}^2$, \[d((x_1,y_1),(x_2,y_2)) := \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.\] A set $S\subseteq\mathbb{R}^2$ is a rational distance set if \[\forall p\neq q\in S,\quad d(p,q)\in\mathbb{Q}.\] A set $S\subseteq\mathbb{R}^2$ is dense if its closure is all of $\mathbb{R}^2$, equivalently \[\forall x\in\mathbb{R}^2\ \forall \varepsilon>0\ \exists s\in S\text{ with } d(x,s)<\varepsilon.\]

Problem #212 (existence form). Does there exist a dense set $S\subseteq\mathbb{R}^2$ such that $S$ is a rational distance set?

2. Quick Literature/Context Check

The prompt itself indicates that this is the classic Erdős--Ulam problem: whether a dense rational distance set exists in the plane. It also indicates conditional negative results (e.g. assuming Bombieri--Lang), and unconditional structural results for rational distance sets lying on algebraic curves. As stated in the prompt, the problem is widely believed to be open.

3. Attack Plan

Proof approach (show existence). Try to build a countable dense set $S$ by an explicit inductive construction:

• Start with a finite rational-distance configuration $S_0$.
• At stage $k$, add a new point $p_k$ very close to a prescribed target point in a countable dense subset of $\mathbb{R}^2$ while forcing all new distances to land in $\mathbb{Q}$.
• The key subproblem: given an existing finite rational-distance set $F$ and a target location $x$, find a point $p$ arbitrarily close to $x$ such that $d(p,f)\in\mathbb{Q}$ for every $f\in F$.

This "local extension" step is the main obstacle; it becomes a Diophantine approximation problem on intersections of many circles.

Disproof approach (show nonexistence). Try to show that any rational distance set $S$ must be contained in a "thin" subset of the plane (e.g. a countable union of curves of empty interior), or must fail to intersect some open ball. A natural route is:

• Prove strong structural constraints on large finite rational-distance sets.
• Upgrade these constraints to infinite sets via compactness/limiting arguments.

The prompt suggests that deep Diophantine geometry methods might be relevant here.

4. Work

I cannot resolve existence/nonexistence in $\mathbb{R}^2$, but I can fully prove some basic structural facts and give explicit dense rational-distance sets on 1-dimensional subsets (lines/circles). These serve as sanity checks and clarify what is and is not ruled out by elementary arguments.

Lemma 4.1 (Countability). Every rational distance set $S\subseteq\mathbb{R}^2$ is at most countable.

Proof. If $|S|\le 1$ there is nothing to prove. Otherwise pick distinct points $A,B\in S$. For each $P\in S$ define \[\Phi(P):=\left(d(P,A),\ d(P,B)\right)\in \mathbb{Q}_{\ge 0}^2. \] Fix any pair $(r,s)\in\mathbb{Q}_{\ge 0}^2$. The fiber \[\Phi^{-1}(r,s)=\{P\in\mathbb{R}^2: d(P,A)=r\text{ and } d(P,B)=s\} \] is the intersection of the two circles \[C_A(r):=\{P: d(P,A)=r\},\qquad C_B(s):=\{P: d(P,B)=s\}. \] Because $A\neq B$, the circles $C_A(r)$ and $C_B(s)$ cannot coincide (distinct centers), and two distinct circles in the Euclidean plane intersect in at most two points. Hence $|\Phi^{-1}(r,s)|\le 2$ for every $(r,s)$. Since $\mathbb{Q}_{\ge 0}^2$ is countable and $S=\bigcup_{(r,s)\in\Phi(S)}\Phi^{-1}(r,s)$, it follows that $S$ is a countable union of finite sets, hence countable. $\square$

Remark. Countability alone does not rule out density, since there exist countable dense subsets of $\mathbb{R}^2$ (e.g. $\mathbb{Q}^2$). So Lemma 4.1 does not resolve #212.

Proposition 4.2 (Dense rational-distance set on a line). The set $L:=\mathbb{Q}\times\{0\}\subset\mathbb{R}^2$ is dense in the $x$-axis and satisfies $d(p,q)\in\mathbb{Q}$ for all $p,q\in L$.

Proof. Density in the $x$-axis is standard: for any real $x$ and $\varepsilon>0$ there exists $q\in\mathbb{Q}$ with $|q-x|<\varepsilon$. For $p=(q_1,0), q=(q_2,0)$ we have $d(p,q)=|q_1-q_2|\in\mathbb{Q}$. $\square$

Proposition 4.3 (Dense rational-distance set on the unit circle). There exists a countable set $S\subseteq\{(x,y):x^2+y^2=1\}$ that is dense on the unit circle and satisfies $d(p,q)\in\mathbb{Q}$ for all distinct $p,q\in S$.

Construction. For $t\in\mathbb{Q}$ define \[a(t):=\frac{1-t^2}{1+t^2},\qquad b(t):=\frac{2t}{1+t^2}. \] Note that $a(t),b(t)\in\mathbb{Q}$ and \[a(t)^2+b(t)^2=\frac{(1-t^2)^2+(2t)^2}{(1+t^2)^2}=\frac{(1+t^2)^2}{(1+t^2)^2}=1. \] Now define \[p(t):=\left(a(t)^2-b(t)^2,\ 2a(t)b(t)\right)\in\mathbb{R}^2. \] Let $S:=\{p(t): t\in\mathbb{Q}\}$.

Proof that $S$ lies on the unit circle. Let $a=a(t)$ and $b=b(t)$. Then \[(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2=1, \] so $p(t)$ lies on $x^2+y^2=1$.

Proof that pairwise distances are rational. Take $t,u\in\mathbb{Q}$ and write $a_t=a(t)$, $b_t=b(t)$, $a_u=a(u)$, $b_u=b(u)$. Consider the angles $\theta,\varphi$ with \[\cos(\theta/2)=a_t,\ \sin(\theta/2)=b_t,\qquad \cos(\varphi/2)=a_u,\ \sin(\varphi/2)=b_u. \] Then $p(t)=(\cos\theta,\sin\theta)$ and $p(u)=(\cos\varphi,\sin\varphi)$, and the chord length formula gives \[d\left(p(t),p(u)\right)=2\left|\sin\left(\frac{\theta-\varphi}{2}\right)\right|. \] Using the sine difference identity, \[\sin\left(\frac{\theta-\varphi}{2}\right)=\sin(\theta/2)\cos(\varphi/2)-\cos(\theta/2)\sin(\varphi/2)=b_t a_u-a_t b_u\in\mathbb{Q}, \] so $d(p(t),p(u))\in\mathbb{Q}$.

Proof that $S$ is dense on the unit circle. The map $t\mapsto (a(t),b(t))$ parametrizes the unit circle minus the point $(-1,0)$, and it is continuous as a function of real $t$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ and the parametrization is continuous, the image $\{(a(t),b(t)):t\in\mathbb{Q}\}$ is dense in the circle. The map $(a,b)\mapsto (a^2-b^2,2ab)$ is continuous and is exactly the angle-doubling map on the circle; the image of a dense subset under a continuous map is dense. Hence $S=\{p(t):t\in\mathbb{Q}\}$ is dense on the unit circle. $\square$

Sticking point for #212. The constructions above are dense on 1-dimensional sets (a line or a circle). Extending these to a set dense in all of $\mathbb{R}^2$ requires controlling rationality of distances between points lying near many different directions/locations simultaneously. Lemma 4.1 shows we must build a countable dense set, but gives no mechanism to enforce the rationality constraints across the entire set.

5. Verification

• Lemma 4.1: the only geometric input is that two distinct circles intersect in $\le 2$ points; this is standard and covers all cases because distinct centers prevent coincident circles.
• Proposition 4.3: all algebraic identities were explicitly expanded; the distance computation reduces to a rational expression via a trigonometric identity in which the relevant sine/cosine values are rational by construction.
• Density: used only continuity and density of $\mathbb{Q}$ in $\mathbb{R}$.

6. FINAL

UNRESOLVED.

1. (i) Strongest fully proved partial result obtained here:
   • Every rational distance set in $\mathbb{R}^2$ is at most countable (Lemma 4.1).
   • There exist countable dense rational-distance sets on a line (Proposition 4.2) and on the unit circle (Proposition 4.3).
2. (ii) First gap blocking completion: There is no known method (in this write-up) to perform the "local extension step": given a finite rational-distance set $F\subset\mathbb{R}^2$ and a target point $x$, construct a point $p$ arbitrarily close to $x$ with $d(p,f)\in\mathbb{Q}$ for all $f\in F$. Without such a step, one cannot inductively build a dense planar set.
3. (iii) Most promising next moves:
   • Develop an extension lemma for finite rational-distance sets using arithmetic geometry of the configuration space cut out by multiple circle equations.
   • Prove an unconditional "finite union of curves" (or other thin-structure) theorem for infinite rational distance sets, strong enough to contradict density.
   • Explore constructions based on group laws on conics/elliptic curves that might allow adding points while controlling many distances at once.
4. (iv) Conjectured final answer and minimal counterexample shape: I conjecture the answer is NO (no dense rational-distance set in $\mathbb{R}^2$). A hypothetical minimal counterexample would necessarily be a countable dense set with highly rigid algebraic structure, likely organized by an underlying algebraic group action that forces all interpoint distances into $\mathbb{Q}$.

7. Completion Estimate

15%. The work here establishes basic necessary structure (countability) and gives dense examples on 1-dimensional sets, but it does not substantially bridge the core difficulty of enforcing rationality of all distances in a set dense in two dimensions.