The Euclidean plane cannot be colored with five colors

The explanation

Color every point of a flat plane, using a small set of colors. There is one rule: two points exactly one unit apart must get different colors. How few colors can work? This is the Hadwiger–Nelson problem, and the answer is called the chromatic number of the plane.

Seven colors are enough. Tile the plane with hexagons whose corners sit 0.4 units from their centers. Two points in one hexagon are at most 0.8 apart, so they never clash. Give the tiles seven colors in a repeating pattern, so that same-colored tiles stay far apart. Points in two different same-colored tiles are then more than 1.03 units apart. No unit pair shares a color.

Three colors are not enough. The Moser spindle, from 1961, is a set of seven points with eleven unit-distance pairs among them. Checking all 2,187 ways to color it with three colors shows that every one breaks the rule. A spindle needs four. Edward Nelson posed the problem in 1950 and found the lower bound of four. John Isbell found the seven.

For almost 70 years the answer sat somewhere between four and seven. In 2018, Aubrey de Grey built a finite set of points that cannot be colored with four colors. His corrected example has 1,581 points. Later work shrank such examples to 553 points, then 509. The lower bound became five. Separately, in 1981, K. J. Falconer had shown that tame colorings need at least five colors. A coloring is tame here when each color covers a region with a well-defined area.

This paper proves that five colors never work, for any coloring at allSource 1. Together with the hexagon tiling, the answer is now six or sevenSource 1. The paper does not decide between themSource 2.

The single difficulty is that a coloring can be wildSource 3. Its color classes need not have any area, so the tools that worked for tame colorings have nothing to grip. The paper removes this in two steps. First it shows that if some coloring with a given number of colors exists, a tame one exists tooSource 4. The tame version may break the rule, but only on a negligible set of pairsSource 5. It is built by averaging the wild coloring over many shifts and rotations of the planeSource 6. Then the paper shows that no tame five-coloring exists, even allowing those negligible exceptions.

That second step tracks where colors meet. Around one well-chosen point, the paper finds connected boundary pieces between pairs of colors. Each piece bars its two colors from a band about one unit away from it. With five colors, these forbidden zones run into each other. In the last case left, some region uses only three colors. The paper places a whole Moser spindle inside itSource 7. Three colors cannot color a spindle, which gives the contradiction.

The proof gives no picture to check by hand. A classical compactness theorem from 1951 says some finite set of points must already defeat five colors. The paper does not exhibit such a set, so its size is unknownSource 8. The earlier five-color bound, by contrast, came with explicit point sets.

How sure is this? A computer proof checker called Lean has checked formal versions of both statements. No coloring with five colors exists, and a seven-color one does. The five-color statement covers arbitrary colorings, with no smoothness assumed. No specialist review of the paper has been published.

What it doesn't show. This does not decide whether the answer is six or seven.