Partition Regularity of Pythagorean Triples

Partition Regularity of Pythagorean Triples

Definitions and mathematical statement

A finite colouring of the positive integers is a function $c:{\mathbb{Z}}_{\ge1}\to\{1,\ldots,r\}$ for some finite $r\ge1$. The Pythagorean equation is partition regular if \[ \forall r\ \forall c\ \exists x,y,z\in{\mathbb{Z}}_{\ge1}, \qquad x^2+y^2=z^2, \qquad c(x)=c(y)=c(z). \] Determine whether this assertion holds. The triple need not be primitive, so a common divisor of its entries is allowed. Positivity excludes zero solutions. The number of colours is arbitrary and fixed for each colouring, not just two; a theorem for one specific number of colours does not settle the universal finite-colouring statement.

Formulation and concept references: [S1].

Short English statement

Does every finite colouring of the positive integers contain a single-coloured Pythagorean triple?

Research attempt

Scope and primary-source audit (27 September 2026).

All three coordinates must have the same color, and the number of colors is arbitrary. The primary survey [S1] distinguishes this from its results about two coordinates and from results for special multiplicative colorings. Its checked text, carrying an August 2026 document date, still states the all-three-coordinate problem as a challenge. The two-color theorem of Heule, Kullmann, and Marek [E1] is a separate established computational result. Its large certificate is not claimed to have been reproduced here.

The following work proves partition regularity for an explicit class of colorings described by finitely many congruences, valuations, and leading local digits. It also proves the exact finite-obstruction reformulation and carefully tracks how parametrization interacts with color. These are partial results and diagnostics; no reduction of arbitrary colorings to the special class is supplied.

Positivity, distinctness, and the primitive reduction.

In a positive Pythagorean triple, $z>x,y$. Also $x\ne y$: otherwise $z^2=2x^2$ would make $\sqrt2=z/x$ rational, which is impossible by the usual parity descent. Thus the statement already forces three distinct positive integers, even though distinctness was not separately imposed.

Dividing a triple by its common divisor gives a primitive triple. In such a triple the three entries are pairwise coprime: a prime dividing two divides the third by the equation. The legs cannot both be odd, since squares modulo four are zero or one; and they cannot both be even by primitivity. Label the odd leg $x$ and the even leg $y$. Then $z$ is odd, and \[ \frac{z+x}{2}\frac{z-x}{2}=\left(\frac y2\right)^2. \] The two factors on the left are coprime: a common divisor divides their sum $z$ and difference $x$. A product of two coprime positive integers is a square only if each factor is a square. Thus \begin{equation}\label{wsref-0-61-33-38-31-3a-70-61-72-61-6d} x=m^2-n^2,\qquad y=2mn,\qquad z=m^2+n^2, \qquad m>n\geq1,\quad\gcd(m,n)=1, \end{equation} with opposite parities for $m,n$. Conversely these conditions give a primitive positive triple; the identity is immediate and the same divisibility argument checks primitivity. Every positive triple is a positive integer multiple of such a triple, up to interchanging the legs.

This is an arithmetic classification, not a color-preserving reduction. Dividing all three entries by a common divisor can change their colors in unrelated ways. For example the parity coloring has no monochromatic primitive triple, but $(6,8,10)$ is monochromatic. Requiring primitiveness would therefore change the problem to a false statement even for two colors. This also explains why the common scaling parameter cannot be suppressed in a proof using (a381:param).

The easiest complete coloring class.

Suppose $c$ is eventually periodic: for some $M,N_0$, its value on integers at least $N_0$ depends only on the residue modulo $M$. Choose $t$ to be a sufficiently large multiple of $M$. Then $3t,4t,5t$ all exceed $N_0$, all have residue zero modulo $M$, and satisfy the equation. Hence every eventually periodic finite coloring contains a monochromatic triple.

In particular, no coloring solely by finitely many residues can disprove the conjecture. Refining a residue coloring by finitely many other invariants is a different matter; those other invariants need not be preserved by multiplication by $M$. The next construction handles a specific and fairly broad family of such refinements.

Local data used by a coloring.

For a prime $q$ and a positive integer $a$, let $v_q(a)$ be the exponent of $q$ in $a$, and let \[ u_q(a)=a/q^{v_q(a)}. \] Thus $u_q(a)$ is an integer prime to $q$. Fix a finite set $S$ of primes and positive integers $r_q,k_q$ for $q\in S$. Consider the finite-valued data \begin{equation}\label{wsref-0-61-33-38-31-3a-6c-6f-63-61-6c-64-61-74-61} \left(v_q(a)\bmod r_q,\quad u_q(a)\bmod q^{k_q}\right)_{q\in S}. \end{equation} We first construct a primitive triple for which all three entries have valuation zero modulo $r_q$ and unit part one modulo $q^{k_q}$, simultaneously for every $q\in S$. The valuations themselves need not be zero; only the specified residues are zero.

Lemma 381.1: a simultaneous local construction.

There are arbitrarily large integers $n$ for which the triple \begin{equation}\label{wsref-0-61-33-38-31-3a-63-6f-6e-73-65-63-75-74-69-76-65} X=2n+1,\qquad Y=2n(n+1),\qquad Z=2n^2+2n+1 \end{equation} has the identical local data just specified. The triple is positive, primitive, and satisfies $X^2+Y^2=Z^2$.

Proof. For each odd $q\in S$, choose $e_q\geq k_q$ divisible by $r_q$, and impose \[ n\equiv q^{e_q}2^{-1}\pmod{q^{e_q+k_q}}, \] where $2^{-1}$ denotes an inverse modulo $q^{k_q}$ in the multiplier on the right. This forces $v_q(n)=e_q$ and $2(n/q^{e_q})\equiv1\pmod{q^{k_q}}$. Since $n\equiv0\pmod{q^{k_q}}$, we get \[ X\equiv Z\equiv1\pmod{q^{k_q}},\qquad v_q(Y)=e_q,\qquad u_q(Y)=2(n/q^{e_q})(n+1)\equiv1\pmod{q^{k_q}}. \] Thus the desired data agree at every odd prime in $S$.

If $2\in S$, choose $e_2\geq\max\{1,k_2\}$ with $e_2+1$ divisible by $r_2$, and impose \[ n\equiv2^{e_2}\pmod{2^{e_2+k_2}}. \] Now $n$ is even and $n+1$ is odd. We have $X\equiv Z\equiv1\pmod{2^{k_2}}$, while \[ v_2(Y)=e_2+1,\qquad u_2(Y)=(n/2^{e_2})(n+1)\equiv1\pmod{2^{k_2}}. \] Again the data are correct. All prescribed moduli are powers of different primes, so the Chinese remainder theorem supplies a progression of solutions $n$, with arbitrarily large positive members. If $S$ is empty no congruences are needed.

Finally (a381:consecutive) is (a381:param) with $m=n+1$. These parameters are coprime and have opposite parity, so the triple is primitive. This also verifies the identity and positivity independently of the congruences.$\square$

Proposition 381.2: a complete finite-local-data subcase.

Suppose that beyond a fixed threshold, $c(a)$ is any function of $a\bmod M$ together with the finitely many data in (a381:localdata). Then $c$ has a monochromatic Pythagorean triple.

Proof. Choose a triple $(X,Y,Z)$ from Lemma 381.1 and multiply it by $M$. All three residues modulo $M$ become zero. Multiplication by the same integer preserves equality of the specified local data: valuations add, and \[ u_q(Ma)=u_q(M)u_q(a) \] as exact integers. Thus their valuations modulo $r_q$ remain equal and their unit residues remain equal. Choose $n$ sufficiently large that $MX,MY,MZ$ all exceed the eventual threshold. Their colors coincide, and homogeneity preserves the Pythagorean equation.$\square$

This includes arbitrary finite products and refinements of the listed invariants, not just one valuation coloring. For example coloring by $v_2(a)$ modulo a fixed integer cannot be an obstruction: the even leg in the construction has a positive valuation divisible by that integer, while the other two valuations are zero. Coloring by the exact valuation takes infinitely many colors and is not an admissible counterexample. Truncating exact valuations to finitely many values changes that situation again, since sufficiently large common scaling makes all three truncated valuations agree.

The proposition is not the much broader theorem for finite-valued multiplicative functions mentioned in [S1]. Nor is every finite coloring determined by finitely many local invariants. This proof supplies no approximation theorem allowing an arbitrary coloring to be replaced by such a coloring without losing the required triple.

Lemma 381.3: the exact finite-obstruction equivalence.

For a fixed number $r$ of colors, the infinite assertion is equivalent to existence of an integer $N(r)$ such that every $r$-coloring of $\{1,\ldots,N(r)\}$ has a monochromatic Pythagorean triple within that interval.

Proof. A finite obstruction immediately works inside every infinite coloring. Conversely suppose that for every $N$ there is an avoiding coloring of $[N]$. Form a tree whose level $N$ consists of these avoiding colorings, with edges given by restriction. There are finitely many extensions of each coloring, at most $r$, and arbitrarily high nonempty levels. Starting at the empty coloring, choose a child having extensions to arbitrarily high levels; such a child exists because a finite union of bounded-height subtrees has bounded height. Repeat. The resulting infinite compatible sequence defines a coloring of all positive integers. Any alleged monochromatic triple occurs in a finite prefix, contradicting avoidance there. This proves the contrapositive.$\square$

The argument is qualitative compactness and gives no bound on $N(r)$. Exhaustive finite search is therefore a semidecision procedure for the positive statement at one fixed $r$: if an obstruction exists, a sufficiently long search can find it. Failure to find one at every tested size does not produce an infinite countercoloring, because a chosen finite coloring may have no extension to the next size.

For $r=1$, the triple $(3,4,5)$ gives $N(1)=5$. For $r=2$, the established computational result [E1] supplies the two-color conclusion. This does not imply the result for $r=3$: merging three colors into two can produce a triple whose coordinates occupy two different original colors that were merged. The implication from a larger number of colors to a smaller one is valid; reversing it is not.

The hypergraph and constraint formulation.

For a finite interval $[N]$, construct a hypergraph with one three-element edge for each positive Pythagorean triple in the interval. An avoiding coloring is exactly a coloring with no monochromatic edge. For two colors represented by Boolean variables $b_a$, each edge $\{x,y,z\}$ gives the two clauses \[ (b_x\vee b_y\vee b_z)\quad\text{and}\quad (\neg b_x\vee\neg b_y\vee\neg b_z). \] The first forbids all three false and the second forbids all three true. This is an exact encoding, including imprimitive triples. Generating only primitive triples would miss constraints and would falsely permit a parity coloring at every finite size.

A satisfying assignment is easy to check by evaluating every generated edge. An unsatisfiability claim is different: it requires exhaustive verification, a checked certificate, or a proof of completeness of the search. Merely timing out or observing that a greedy rule gets stuck proves neither unsatisfiability nor partition regularity. The use of computer certification in [E1] is consequently substantive, rather than a cosmetic numerical illustration.

Two tempting linear arguments fail at a specific step.

Schur's theorem gives same-colored $a,b,a+b$ for any finite coloring of the integers. If it is applied to the derived coloring $d(t)=c(t^2)$, it gives equal colors for $a^2,b^2,(a+b)^2$. But $a^2+b^2\ne(a+b)^2$ for positive $a,b$, because of the nonzero term $2ab$. Applying Schur's theorem directly to the set of squares would require a theorem about that restricted domain, not Schur's theorem as stated.

Similarly, choosing $m,n$ of the same color in (a381:param) says nothing about the colors of $m^2-n^2$, $2mn$, and $m^2+n^2$. The parametrization identifies the values that must be compared; it does not make the color map compatible with addition or multiplication. The CRT proof works precisely because it imposes explicit compatibility data on that color map. Its missing extension is control of colors with no such finite arithmetic description.

Exact diagnostics and outcome.

The companion calculation generates triples by integer square-root tests, independently checks the Euclidean identities and primitivity, and constructs CRT witnesses for several finite lists of primes and valuation/unit moduli. It verifies equality of every local datum before and after common scaling. It also finds a two-color avoiding assignment on $[100]$ by a complete finite backtracking routine and then checks that assignment directly against all triples. This small positive assignment is not a rerun of the large two-color obstruction theorem.

UNRESOLVED. The attempt proves the full conclusion for the finite-local-data coloring class in Proposition 381.2 and an exact compactness reduction for each fixed $r$. The first remaining gap is a way to force simultaneous equality of all three colors for an unrestricted finite coloring, beyond its residue and valuation data. Neither the parametrization, the proved two-color theorem, nor the pair-coloring theorems in the source supply that missing all-colors statement here.

Sources

• S1 Partition regularity of homogeneous quadratics: Current trends and challenges. Submitted scholarly source, accessed 2026-08-31. https://arxiv.org/abs/2411.17523. Formulation source.
• E1 M. J. H. Heule, O. Kullmann, and V. W. Marek, Solving and Verifying the Boolean Pythagorean Triples Problem via Cube-and-Conquer, SAT 2016, LNCS 9710, 228--245; arXiv:1605.00723. https://arxiv.org/abs/1605.00723. Primary two-color result checked 27 September 2026. Its large unsatisfiability certificate is attributed, not rerun in this attempt.