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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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Schur's theorem: every finite colouring of a sufficiently long positive initial interval {1,,N}\{1,\ldots,N\} has positive monochromatic x,y,zx,y,z with x+y=zx+y=z

Statement

For every positive number cc of colours there is a positive natural NN such that every cc-colouring of {1,,N}\{1,\ldots,N\} has positive x,y,zx,y,z of one colour satisfying x+y=zx+y=z. The variables need not be distinct. Natural order is that of The natural numbers N\mathbb{N} (von Neumann) and Order on the natural numbers, and the proof uses the pair-colouring convention of Finite colourings of kk-element subsets, monochromatic sets, and the arrow notations N(s,t)2N\to(s,t)^2 and N(r)ckN\to(r)^k_c.

Facts & Assumptions

Given: A positive number cc of colours and a colouring of a sufficiently long positive initial interval.

[L1]

For all positive s,ts,t, (s+t2s1)(s,t)2\binom{s+t-2}{s-1}\to(s,t)^2 (Finite graph Ramsey theorem: (s+t2s1)(s,t)2\binom{s+t-2}{s-1}\to(s,t)^2 for all positive s,ts,t).

Proof

technique · direct
1.1

Iterating [L1] gives a finite MM such that every cc-colouring of the pairs of an MM-element set has a monochromatic triangle: separate one colour from the remaining colours, use [L1] with target 33 for the first colour and with a recursively chosen target for the others, and continue through the finite colour list.

L1
2.1

Colour the edge {i,j}\{i,j\} of the ordered vertex set {0,,M1}\{0,\ldots,M-1\}, with i<ji<j, by the given colour of the positive difference jij-i. Step 1.1 gives a monochromatic triangle i<j<ki<j<k.

step 1.1
3.1

Put x=jix=j-i, y=kjy=k-j and z=kiz=k-i. These are positive, the edge colouring says they have one original colour, and arithmetic gives x+y=zx+y=z. Taking N=M1N=M-1 contains all three differences.

step 2.1algebra

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 37 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources