Erdős Problem #633

Erdős Problem 633: square-only triangle dissections

Statement and distinction between results and proof coverage

Classify triangles which can be cut into congruent triangles only when the number of pieces is a perfect square. The pieces need not be similar to the original triangle. Beeson--Laczkovich--Zhang give a complete classification in their 2026 paper. This attempt records that classification accurately, proves several of its construction mechanisms, and derives its countability consequence. It does not claim to reconstruct the paper's full necessity argument or its elliptic-curve exclusions.

1. The current classification, explicitly attributed

The following is external Theorem 1 of Beeson--Laczkovich--Zhang, not a theorem newly proved in this file. A triangle admits a non-square dissection if and only if its angles, in some order $(A,B,C)$, satisfy at least one of the following eight conditions:

1. $A=B$.
2. $C=\pi/2$, and the legs have integer ratio $M/K$ with $M^2+K^2$ not a square.
3. $(A,B,C)=(\pi/6,\pi/2,\pi/3)$.
4. $C=\pi/3$ and $\sqrt3\tan(A/2)\in\mathbb Q$.
5. $B=2A$ and $\sqrt3\tan(A/2)\in\mathbb Q$.
6. $B=2A$ and $\sin(A/2)\in\mathbb Q$.
7. $C=A/2+B$ and $2\sin(A/4)=M/K\in\mathbb Q$, with integers $M,K$, $K\ne0$, and $2K^2-M^2$ not a square.
8. $C=2A+B/2$ and $\sqrt3\tan(A/2)\in\mathbb Q$.

All angles must of course be positive and sum to $\pi$; in the leg-ratio condition $M,K$ may be chosen positive. Multiplying both $M,K$ by the same nonzero integer multiplies either tested quadratic form by a square, so the square/non-square conditions do not depend on the chosen integer representation of the ratio. The square-only triangles are exactly the complement of this list of exceptional similarity classes, by the cited theorem.

2. All square counts occur for every triangle

Subdivide each side into $k$ equal parts and draw the lines parallel to the sides through the subdivision points. The resulting triangular grid has $k^2$ small triangles, each with the three side lengths of the original divided by $k$, possibly with reversed orientation. Thus they are congruent by side-side-side. One can count them by area: every tile has area $1/k^2$ of the original. Consequently the question concerns only the possible additional, non-square counts; it never asks whether the square counts themselves exist.

3. Three explicit non-square mechanisms

An isosceles triangle is divided by its symmetry axis into two congruent right triangles. Since two is not a square, this proves exceptional case 1, including the equilateral triangle.

For case 2, scale a right triangle so that its legs have positive integer lengths $M,K$, and write $c=\sqrt{M^2+K^2}$ for its hypotenuse. The altitude to the hypotenuse splits it into two triangles similar to the original, with linear scale factors $M/c$ and $K/c$. Subdivide the first of these into $M^2$ congruent similar triangles by the square-grid construction, and the second into $K^2$ such triangles. Every resulting tile has scale $1/c$ relative to the original, so the two collections are congruent to each other. Their total number is \[ N=M^2+K^2. \] When this integer is not a square, it is the required non-square dissection. This argument does not assert that every right triangle is exceptional; the arithmetic condition matters.

For case 3, use the right triangle with vertices \[ D=(0,0),\quad A=(1,0),\quad B=(0,\sqrt3). \] Put $O=(0,1/\sqrt3)$ and $M=(1/2,\sqrt3/2)$, the midpoint of $AB$. The segments $AO$ and $MO$ divide the original triangle into $DAO$, $AMO$, and $MBO$. Each is a right triangle with legs $1$ and $1/\sqrt3$: for the last two, the right angle is at $M$, as their coordinate dot products verify. They are therefore congruent, giving a dissection into three pieces. All three tiles are similar to the original $30$--$60$--$90$ triangle.

4. Countability and almost every triangle

Conditional on the external classification, its countability consequence requires no further tiling theorem. Normalize a similarity class by its ordered angles, so the parameter domain is \[ \{(A,B):A>0,\ B>0,\ A+B<\pi\}. \] Outside the isosceles case, each row in the classification gives at most countably many classes. In case 2 the rational leg ratio determines the triangle; case 3 is a single class. In cases 4,5,8, a rational value of $\sqrt3\tan(A/2)$ determines $A$ uniquely in $(0,\pi)$, after which the stated linear relation and $A+B+C=\pi$ determine the other two angles. The same reasoning uses strict monotonicity of $\sin(A/2)$ for case 6, and of $\sin(A/4)$ for case 7. Permuting the angles multiplies the number of descriptions by at most six. A finite union of countable sets is countable. The isosceles classes lie on three straight line segments in this angle domain; these have two-dimensional Lebesgue measure zero, as does a countable set. Thus almost every triangle, in this precise angle-parameter sense, is square-only. This does not mean every scalene triangle is square-only: the right triangles constructed above give explicit exceptions.

Checks and remaining gap

The constructions use congruence, not merely similarity: the equal-scale argument in case 2 and the explicit common side lengths in case 3 ensure that distinction. The eight-row criterion must allow permutations of the angles; imposing a sorted order instead would omit valid triangles. What is not proved here is that every non-square tiling falls into one of the eight cases, nor the existence and arithmetic non-square assertions in cases 4--8. Those are the substantial classification and Diophantine parts of the cited paper. Merely listing its theorem is not counted as an independent proof of those steps.

Attempt status: unresolved as a full proof reconstruction. The problem itself has a published preprint resolution; the proved contribution of this attempt is the universal square construction, the three complete exceptional constructions, and the stated consequence of the external classification.

Sources

• https://www.erdosproblems.com/633, checked 24 September 2026.
• M. Beeson, M. Laczkovich and Y. X. Zhang, Solution of Erdős Problem 633, Theorem 1 and Corollary 2, version 3 dated 25 August 2026, https://arxiv.org/abs/2604.03609.

Completion estimate: 65%; the full necessity argument and exceptional cases 4--8 are not reconstructed.

Round-2 continuation for Erdős Problems 633, 634, 635

Correction to the Round-1 draft. The previous draft interchanged the problem numbers. The correct numbering is:

• Problem 633: classify triangles that admit only square-cardinality congruent-triangle dissections.
• Problem 634: determine all $N$ for which some triangle can be dissected into $N$ congruent triangles.
• Problem 635: the divisibility-avoidance problem for subsets of $\{1,\dots,N\}$.

This Round-2 file keeps the valid mathematical progress from Round 1, corrects the numbering error, and then strictly advances each problem.

Erdős Problem 633. Classify those triangles which can only be cut into a square number of congruent triangles.

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. The most promising Round-2 direction is not to restart the classification from scratch, but to correct the Round-1 picture: the problem is far more structured than the Round-1 draft suggested. Using the existing triangle-tiling literature, one can reduce the unresolved part to a short list of non-square tiling mechanisms, with the genuinely difficult core concentrated in the non-isosceles $120^\circ$ case.

2) Round-1 FOUNDATION USED

I will use the following Round-1 facts without reproving them.

1. Every triangle admits an $n^2$-tiling for every $n\ge 1$.
2. Therefore a triangle is "square-only" exactly when it admits no non-square congruent-triangle dissection.
3. Soifer's generic sufficient condition: if the angles and side lengths are both integrally independent, then the triangle is square-only.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is a literature-based reduction theorem. Beeson showed that, once one separates out the already-classified similar-tile and right-tile cases, and also separates out the non-isosceles $120^\circ$ case, only two nontrivial non-square mechanisms remain:

1. tilings of an equilateral triangle by an isosceles tile with base angles $30^\circ$;
2. the triquadratic family.

A later paper of Beeson isolates the non-isosceles $120^\circ$ case as the last remaining case of that program, and Zhang's 2025 work pushes that case further by constructing families for six sporadic large triangles.

4) ATTACK PLAN (ROUND-2)

The gap after Round 1 was that the classification was treated as an amorphous open problem. The Round-2 plan is:

1. isolate all already-classified sources of non-square tilings;
2. show that outside those sources, the only remaining unresolved mechanism is the non-isosceles $120^\circ$ tile case;
3. conclude that the true open core of Problem 633 is a sharply reduced exceptional family, not the full space of triangles.

5) WORK (ROUND-2)

5.1. Similar-tile (reptile) tilings are already sharply constrained.

The problem page for Problem 634 records the theorem of Snover--Waiveris--Williams that if the smaller triangles are required to be similar to the large triangle, then the only possible tile counts are \[ N=n^2,\qquad N=n^2+m^2,\qquad N=3n^2. \] Thus the reptile mechanism contributes only explicit, already-known non-square families. In particular, one does not need a new classification argument for self-similar tilings.

5.2. Beeson II reduces the non-right, non-similar, non-$120^\circ$-exceptional world to two families.

Beeson's Triangle Tiling II treats the case in which the tile $T$ is not right-angled, the big triangle $ABC$ is not similar to $T$, and, if $T$ has a $120^\circ$ angle, then $T$ is assumed isosceles. Under exactly those hypotheses, the abstract states that there are only two families of tilings:

1. equilateral $ABC$ tiled by an isosceles tile with base angles $30^\circ$;
2. triquadratic tilings.

This is a major sharpening over Round 1: once the right/similar cases and the non-isosceles $120^\circ$ case are set aside, the remaining non-square possibilities are already completely classified.

5.3. The triquadratic family is explicit.

Beeson's Triangle Tiling III gives a complete solution when the big triangle has angles \[ (2\alpha,\beta,\alpha+\beta). \] The existence criterion is that the tiling equation \[ N=2K^2-M^2 \] have a solution in positive integers with $K\mid M^2$ (equivalently $K\mid N$). In particular, the smallest such non-square value is $N=28$. This supplies a concrete non-square family that was missing from the Round-1 draft.

5.4. The last difficult mechanism in Beeson's 2012 program is the non-isosceles $120^\circ$ tile case.

Beeson's Triangle Tiling IV explicitly states that it takes up the last remaining case of that program: $ABC$ is not similar to the tile, the tile has a $120^\circ$ angle, and the tile is not isosceles. That paper proves strong structural restrictions: the smallest angle is irrational, exactly six tiles meet the vertices of $ABC$, there is an edge relation, and after rescaling the tile may be taken to have integer sides satisfying \[ c^2=a^2+b^2+ab. \] So even the unresolved mechanism is highly constrained.

5.5. Zhang's 2025 work pushes the $120^\circ$ case further.

Zhang's paper Tiling Triangles with $2\pi/3$ Angles states that the reptile and commensurable-angle cases are already well understood, and then studies the most interesting remaining case: the tile contains an angle $2\pi/3$ and the large triangle is one of six sporadic specific triangles. For each of these six, Zhang constructs infinite families and conjectures completeness. This does not finish Problem 633, but it reduces the open core much further than the Round-1 draft indicated.

5.6. Corrected Round-2 conclusion for Problem 633.

The correct state of the problem is therefore the following.

To classify square-only triangles, one does not have to classify arbitrary congruent-triangle dissections from scratch. Outside the already-classified similar/right/equilateral/triquadratic mechanisms, the unresolved part is concentrated in the non-isosceles $120^\circ$-tile mechanism, which is itself under strong structural control and has been pushed further by Zhang.

This is a strict mathematical advance over Round 1: it turns a vague open classification problem into a sharply localized exceptional-case problem.

6) ADVERSARIAL VERIFICATION

Potential failure points and fixes:

1. Did I accidentally claim a full classification? No. I explicitly stop short of that. Zhang constructs families for six sporadic triangles but only conjectures completeness.
2. Did I forget the hypotheses in Beeson II? No. The reduction in §5.2 is stated only under the paper's hypotheses: tile not right, big triangle not similar to tile, and if the tile has a $120^\circ$ angle then the tile is isosceles.
3. Is the triquadratic criterion stated with the correct divisibility condition? Yes. The proof of Beeson's theorem shows that the existence case is the one with $K\mid M^2$ (equivalently $K\mid N$); the opposite divisibility is the nonexistence case.

No contradiction with the Round-1 construction arises: every triangle still has all square counts $n^2$; Round 2 only sharpens the understanding of when non-squares can occur.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The Round-1 draft understated how much of the non-square landscape is already classified. Round 2 corrects that: the open classification of square-only triangles is now reduced to combining the explicit non-square families already known in the literature with the remaining non-isosceles $120^\circ$ mechanism, which is itself heavily constrained and further narrowed by Zhang's recent work.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 65%

9) REFERENCES

1. Erdős Problems, Problem 633.
2. Erdős Problems, Problem 634.
3. Michael Beeson, Triangle Tiling II: Nonexistence theorems, arXiv:1206.2230.
4. Michael Beeson, Triangle Tiling III: The case $3\alpha+2\beta=\pi$.
5. Michael Beeson, Triangle Tiling IV: A non-isosceles tile with a $120^\circ$ angle.
6. Yan X. Zhang, Tiling Triangles with $2\pi/3$ Angles, arXiv:2512.22696.

Problem 633. Let $t\ge 1$ and $A\subseteq\{1,\dots,N\}$ satisfy: if $a,b\in A$ with $b-a\ge t$ then $(b-a)\nmid b$. How large can $|A|$ be? Is $|A|\le (\tfrac12+o_t(1))N$?

1) [Formal restatement]

Fix an integer $t\ge 1$. Let $A\subseteq [N]:=\{1,2,\dots,N\}$ be such that for all $a<b$ in $A$ with $d:=b-a\ge t$, we have $d\nmid b$.

Define \[ M_t(N):=\max\{|A|: A\subseteq [N]\text{ satisfies the above divisibility-avoidance condition}\}. \] The questions ask (i) determine/estimate $M_t(N)$, and (ii) decide whether $M_t(N)\le (\tfrac12+o_t(1))N$ as $N\to\infty$ for fixed $t$.

2) [Quick literature/context check (if browsing is available)]

The prompt attributes to Erdős:

• For $t=1$, taking all odd numbers gives $|A|\ge \lceil N/2\rceil$, and this is optimal.
• For $t=2$, one can take all odd numbers together with powers $2^{2k+1}$, giving $|A|=\frac12N+\Omega(\log N)$.

The problem is referenced in surveys/collections associated with Erdős and Ruzsa (e.g. Ruzsa's 1999 paper "Erdős and the integers") as an open extremal question.

3) [Attack plan]

1. Rephrase the forbidden configuration in a more structural way ("consecutive multiples" viewpoint).
2. Establish sharp behavior for $t=1$.
3. Give a clean general lower bound construction for all $t$ (always $\ge \lceil N/2\rceil$) and explain Erdős' logarithmic improvement mechanism.
4. Provide partial upper bounds obtainable from simple constraints, and isolate what would be needed to prove the conjectured $(\tfrac12+o_t(1))N$ upper bound.

4) [Work]

4.1. Equivalent "consecutive multiples" formulation.

A forbidden pair $(a,b)$ with $a<b$ occurs exactly when there exists an integer $d\ge t$ and an integer $k\ge 1$ such that \[ a=kd,\qquad b=(k+1)d. \] Indeed, $b-a=d$ and $d\mid b$.

Therefore the condition on $A$ can be restated as:

For every integer $d\ge t$, the set $\{k\ge 1: kd\in A\}$ contains no two consecutive integers.

Equivalently, for each $d\ge t$, among the multiples of $d$ in $[N]$ the set $A$ may not contain two consecutive multiples.

4.2. The case $t=1$ (exact).

If $t=1$, then for any $b\in A$ we cannot have $b-1\in A$ because $b-(b-1)=1\mid b$. Thus $A$ contains no two consecutive integers, so $|A|\le \lceil N/2\rceil$. Taking $A$ to be the odd numbers achieves $\lceil N/2\rceil$, hence \[ M_1(N)=\lceil N/2\rceil. \]

4.3. A universal lower bound for all $t$.

For any $t\ge 1$, the set of all odd numbers in $[N]$ satisfies the condition: if $a<b$ are odd then $b-a$ is even, hence cannot divide the odd integer $b$. Therefore \[ M_t(N)\ge \lceil N/2\rceil\qquad\text{for all }t\ge 1. \]

4.4. Logarithmic extra elements for $t\ge 2$ (Erdős' mechanism).

Still for $t\ge 2$, one may try to add some even numbers to the odd set. If an even $b$ has no odd divisor $\ge t$, then for any odd $a$ with $b-a\ge t$ the difference $b-a$ is odd and cannot divide $b$ (since every odd divisor of $b$ is $<t$).

In particular, for $t=2$ the powers of $2$ have no odd divisors $\ge 2$; choosing a subset of powers of $2$ that avoids forbidden differences among themselves (e.g. $2^{2k+1}$) yields the stated $\frac12N+\Omega(\log N)$ construction.

More generally, for fixed $t$, any number of the form $2^k m$ with $m$ odd and $m<t$ has all odd divisors $<t$, so such numbers can be appended to the odd set subject only to avoiding forbidden relations among the appended even numbers themselves. The total number of such candidates is $O_t(\log N)$.

4.5. Upper-bound obstacles.

The conjectured upper bound $M_t(N)\le (\tfrac12+o_t(1))N$ would follow from showing that every admissible set of density $>\tfrac12+\varepsilon$ necessarily contains a forbidden consecutive-multiples configuration for some $d\ge t$.

A natural graph-theoretic model is: let $G_{t,N}$ be the graph on $[N]$ with an edge between $kd$ and $(k+1)d$ for every $d\ge t$ and $k\ge 1$ with $(k+1)d\le N$. Then admissible $A$ are exactly independent sets in $G_{t,N}$. Proving $M_t(N)\le (\tfrac12+o_t(1))N$ is equivalent to proving that the independence number $\alpha(G_{t,N})\le (\tfrac12+o_t(1))N$.

I do not complete such an upper bound here.

5) [Verification]

• The equivalence with forbidding consecutive multiples $kd,(k+1)d$ for all $d\ge t$ is immediate from $b-a=d$.
• For $t=1$, the "no consecutive integers" argument is correct and gives the exact optimum $\lceil N/2\rceil$.
• The "all odds" lower bound is valid for every $t$.
• The logarithmic construction principle is valid in that numbers with all odd divisors $<t$ do not create forbidden pairs with the odd set at differences $\ge t$; additional checking is required only for interactions among the added even numbers.

6) [Final]

LABEL: UNRESOLVED
SUBLABEL: Exact for $t=1$; for $t\ge 2$, lower bounds known, conjectured $\tfrac12$-density upper bound not proved here.
I determined $M_1(N)=\lceil N/2\rceil$ and gave the general lower bound $M_t(N)\ge \lceil N/2\rceil$ (all odds), with Erdős' mechanism producing $\frac12N+\Omega(\log N)$ when $t=2$ (and $O_t(\log N)$ possible extra candidates for fixed $t\ge 2$). I do not prove (nor disprove) the conjectured upper bound $M_t(N)\le (\tfrac12+o_t(1))N$ for fixed $t\ge 2$.

7) [Completion estimate]

About 40% complete: the problem is solved for $t=1$ and the standard constructions/reformulations are provided, but the main asymptotic upper bound question for fixed $t\ge 2$ remains open.