Square peg problem

Does every continuous loop in the plane that never crosses itself contain four points forming a square of positive size? On a circle, the top, right, bottom and left points give a familiar example. The question includes loops with arbitrarily rough boundaries, without assuming a smooth tangent.[1]

Can every continuous, simple closed planar loop supply four distinct points making a square with positive side length? A Jordan curve is such a loop; no differentiability is assumed, and the square's edges or interior need not stay inside it.

For a unit circle, choose $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$ in that order. All four sides have length $\sqrt{2}$, and neighboring sides meet at right angles. For a general loop, four boundary points must satisfy those length and angle requirements simultaneously. Approximating the loop by smoother ones gives candidate squares, but their sides might approach zero; a limit consisting of one point would not answer the question.[1]

It is proved for many well-behaved loops. Highly irregular continuous loops remain a problem; squares in approximations can shrink away.[1] [2]

The general continuous case remains open. Smooth approximations may yield only a degenerate limiting square, and the June 2026 rhombus results establish a different shape condition.[1] [2]

It would show whether an intuitive geometric pattern survives even in loops far rougher than familiar drawings.[1]

The question separates the topology of a loop from exact metric information: closing without self-crossings says nothing directly about equal distances or perpendicular sides. A positive answer would show that this weak global connection still forces a rigid four-point configuration. The collapsing-square obstacle explains why drawing increasingly accurate smooth copies cannot by itself establish that connection.[1]