Erdős Problem #634

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 634. Find all values $N$ such that there exists a triangle which can be cut into $N$ congruent triangles.

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. The best Round-2 move is to correct the incomplete family list from Round 1. I will show that the known set of attainable $N$ is substantially larger than the Round-1 writeup recorded, by adding (i) the triquadratic family and (ii) a recent theorem of Zhang implying that for every fixed integer $m\ge 1$, all sufficiently large numbers of the form $mn^2$ are attainable.

2) Round-1 FOUNDATION USED

I rely on the following Round-1 material.

1. Every square $n^2$ lies in the attainable set $\mathcal N$.
2. The classical additional families recorded on the problem page are \[ 2n^2,\qquad 3n^2,\qquad 6n^2,\qquad n^2+m^2. \]
3. Beeson showed that $7,11\notin\mathcal N$.

3) NEW INSIGHT / TOOL (ROUND-2)

Two genuinely new ingredients are added.

1. Triquadratic family. Beeson III supplies an additional explicit infinite family, governed by the equation $N=2K^2-M^2$ with a divisibility condition.
2. Zhang's square-multiple theorem. The Erdős problem page now records a theorem implying that for every fixed $m\ge 1$, all sufficiently large values of the form $mn^2$ belong to $\mathcal N$.

These were missing from the Round-1 draft.

4) ATTACK PLAN (ROUND-2)

The gaps after Round 1 were:

1. the known existence side of $\mathcal N$ was incomplete;
2. no attempt was made to extract any corollary from the recent theorem quoted on the problem page.

I will now repair both gaps, while keeping the status honest: the exact determination of $\mathcal N$ is still open.

5) WORK (ROUND-2)

5.1. Classical attainable families.

From the problem page we have \[ n^2,\qquad 2n^2,\qquad 3n^2,\qquad 6n^2,\qquad n^2+m^2\in \mathcal N. \] We also know that $7,11\notin\mathcal N$, and $19$ is still open.

5.2. New family omitted in Round 1: triquadratic values.

Beeson's Triangle Tiling III gives a complete solution for the family of large triangles with angles $(2\alpha,\beta,\alpha+\beta)$. The criterion is: \[ N\in \mathcal N \text{ via such a triangle} \quad\Longleftrightarrow\quad N=2K^2-M^2 \] for some positive integers $K,M$ with $K\mid M^2$ (equivalently $K\mid N$).

This is a real enlargement of the known family list. For example, \[ 28=2\cdot 4^2-2^2 \] with $4\mid 2^2$, so $28\in\mathcal N$. Moreover $28$ is not of any of the classical forms listed in §5.1:

• it is not a square;
• it is not $2n^2$, $3n^2$, or $6n^2$;
• it is not a sum of two squares, since a direct check gives \[ 28-0^2=28,\quad 28-1^2=27,\quad 28-2^2=24,\quad 28-3^2=19,\quad 28-4^2=12,\quad 28-5^2=3, \] and none of these is a square.

Thus Round 1 really was incomplete: $28$ is a new explicit attainable value.

5.3. A strong corollary of the recent theorem on the problem page.

The current problem page records the following theorem: for integers $a\ge b$, if \[ n\ge 3\left\lceil \frac{a^2+b^2+ab-a-b}{ab}\right\rceil, \] then $n^2ab\in\mathcal N$.

Fix now any integer $m\ge 1$ and choose $(a,b)=(m,1)$. Then \[ \frac{a^2+b^2+ab-a-b}{ab} = \frac{m^2+1+m-m-1}{m}=m, \] so the theorem yields the explicit corollary \[ mn^2\in\mathcal N\qquad\text{for every } n\ge 3m. \] This is a substantial strengthening of the Round-1 discussion. In particular:

For every fixed $m\ge 1$, infinitely many multiples of $m$ belong to $\mathcal N$, and in fact every sufficiently large square-multiple $mn^2$ does.

5.4. What remains open.

The exact set $\mathcal N$ is still not known. The small obstructions $7$ and $11$ remain valid, and $19$ is still specifically open. The recent theorem proves large new infinite families, but not a complete characterization.

6) ADVERSARIAL VERIFICATION

Checks on possible weak points:

1. Could $28$ already belong to the old list? No; the direct calculation in §5.2 rules out the sum-of-two-squares possibility, and the other classical forms are obviously impossible.
2. Did the Zhang corollary use the theorem correctly? Yes. Substituting $(a,b)=(m,1)$ gives the threshold $n\ge 3m$ exactly.
3. Did I overclaim a full classification? No. I only enlarge the proven existence side of $\mathcal N$ and keep the exact determination open.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 corrects the existence side of Problem 634 in a mathematically substantive way. Beyond the classical families $n^2,2n^2,3n^2,6n^2,n^2+m^2$, there is also the triquadratic family $N=2K^2-M^2$ with $K\mid M^2$, giving new explicit values such as $28$, and the current problem page records a theorem implying that for every fixed $m\ge 1$, all $mn^2$ with $n\ge 3m$ are attainable. The exact description of $\mathcal N$ remains open.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 60%

9) REFERENCES

1. Erdős Problems, Problem 634.
2. Michael Beeson, Triangle Tiling III: The case $3\alpha+2\beta=\pi$.
3. Yan X. Zhang theorem as quoted on the current Erdős Problems page for Problem 634.

Erdős Problem #634: independent continuation

Date: 23 September 2026. Source versions read in full: 634.tex in gpt_pro_5.4.

1) OBJECTIVE AND NUMBERING

Problem 634 asks which integers occur as the number of congruent triangles tiling a triangle. The supplied file also mentions 633 and 635; those are different problems and are not silently treated as versions of 634. The task here is an arithmetic refinement of the sufficient condition quoted there, not a classification.

2) BALANCING THE FACTORIZATION

Use Zhang's theorem in the form recorded in the official problem discussion: for positive integers $a\ge b$, the number $n^2ab$ is attainable if \[ n\ge3\left\lceil\frac{a^2+b^2+ab-a-b}{ab}\right\rceil. \] For $m\ge1$, define the explicitly computable sufficient threshold \[ T(m)=3\min_{ab=m,\ a\ge b}\left\lceil\frac ab+\frac ba+1-\frac1a-\frac1b\right\rceil. \] Then every $mn^2$ with $n\ge T(m)$ is attainable. The source uses only $(a,b)=(m,1)$ and thus obtains $3m$. Taking the minimum is always at least as good, and can be much better. For instance, $m=35=7\cdot5$ gives the expression $97/35$, hence $T(35)\le9$, compared with the source threshold $105$. For $m=143=13\cdot11$, it gives $409/143$, hence $T(143)\le9$, compared with $429$.

3) UNIFORM BOUND FOR BALANCED FACTORS

If $m=ab$ with $1\le a/b\le R$, then \[ T(m)\le3\lceil R+2\rceil, \] since $b/a\le1$ and the two negative terms can be dropped. Thus the sufficient square multiplier can be bounded independently of $m$ on any family admitting uniformly balanced factors. This follows directly from the quoted theorem; it is not a new tiling construction. More generally, for a proposed count $N$, enumerate every factorization $N=mn^2$ and every $m=ab$. Any one successful inequality certifies attainability. It is unnecessary to restrict to the squarefree decomposition.

4) LIMITATION

Failure of this finite sufficient test proves nothing about nonexistence. In particular it cannot classify small exceptional counts or replace the geometric hypotheses in other tiling theorems. I do not use the source's unrestricted triquadratic equivalence without checking all its geometric side conditions.

7) FINAL

UNRESOLVED. The improvement is an optimized arithmetic certificate for an existing sufficient theorem, not a solution of the classification problem. The theorem was checked against the official record at https://www.erdosproblems.com/634; no independent proof of Zhang's construction 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.

Problem 634. Find all values $n$ such that there exists a triangle which can be cut into $n$ congruent triangles.

1) [Formal restatement]

Determine the set \[ \mathcal{N}:=\{n\in\mathbb{Z}_{\ge 1}: \exists\text{ a triangle }T\text{ that can be partitioned into }n\text{ congruent triangles}\}. \] Equivalently, decide for each $n$ whether at least one triangle admits an $n$-triangle congruent dissection.

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

From the prompt and widely cited tiling constructions:

• $n=k^2$ always lies in $\mathcal{N}$ (by the explicit construction proved in Problem 635).
• Various additional infinite families are known to lie in $\mathcal{N}$ (e.g. $2k^2,3k^2,6k^2$, and $k^2+m^2$ via right-triangle similarity constructions).
• Beeson (as cited in the prompt) shows that $7,11\notin\mathcal{N}$.
• Recent work (e.g. Zhang's 2025/2026 arXiv preprint cited on the Erdős Problems site) studies tilings where the tile has a $2\pi/3$ angle, giving further structure and conjectures, but not a complete classification of $\mathcal{N}$.

3) [Attack plan]

1. Separate the problem into (A) "reptile" (tile similar to the whole) cases and (B) non-similar congruent tilings.
2. Record unconditional constructions that realize large families of $n$.
3. Record unconditional obstructions (e.g. $n=7,11$) and known necessary conditions from angle/edge arithmetic.
4. Explain why closing the gap to an exact description of $\mathcal{N}$ remains open.

4) [Work]

4.1. Constructions giving infinite subfamilies of $\mathcal{N}$.

• Squares. For every $k\ge 1$, there exists a triangle cut into $k^2$ congruent triangles (Problem 635, §4.1).
• Doubles of squares. Using the same parallelogram grid but taking the entire parallelogram rather than half gives a dissection of a parallelogram into $2k^2$ congruent triangles; choosing a triangular half of a suitable parallelogram and refining yields $2k^2\in\mathcal{N}$.
• Sums of two squares (right-triangle similarity). There are standard "reptiling" constructions for right triangles where a larger similar right triangle is formed from a rectangular array of congruent right-triangle tiles, yielding tile counts of the form $k^2+m^2$.
• Multiples $3k^2$ and $6k^2$. Equilateral/60--60--60 and 30--60--90 angle structures allow triangular-lattice style constructions that yield $3k^2$ and $6k^2$ families.

(Here the point is not the uniqueness of the construction, but the existence of triangles realizing these $n$.)

4.2. Obstructions.

The prompt reports that $7,11\notin\mathcal{N}$ (Beeson). Beyond these sporadic nonexistence results, general necessary conditions often come from:

• Angle constraints: each angle of the big triangle is a sum of tile angles at boundary vertices; and the multiset of tile angles around interior vertices must sum to $2\pi$.
• Edge constraints: each side length of the big triangle is a nonnegative integer combination of the tile side lengths, respecting which tile side can lie on which boundary side.

These are powerful enough to rule out some small $n$, but (at present) do not yield a complete characterization of $\mathcal{N}$.

5) [Verification]

• The existence of $k^2\in\mathcal{N}$ is rigorously verified by the explicit $k^2$-tiling construction in Problem 635.
• The additional families listed are consistent with standard tiling constructions; the key unresolved step is proving that no other $n$ occur beyond the known families and sporadic cases, or else producing new constructions for the remaining $n$.

6) [Final]

LABEL: UNRESOLVED
SUBLABEL: Partial progress (recorded constructions/obstructions) + full classification open.
I can certify large infinite families inside $\mathcal{N}$ (in particular all squares) and note known sporadic nonexistence ($7,11$), but I do not obtain a complete description of all $n$ for which some triangle admits an $n$-congruent-triangle dissection.

7) [Completion estimate]

About 25% complete: the square-family is proved constructively and the known landscape is organized, but the "if and only if" characterization of $\mathcal{N}$ is not reached.