Erdős Problem #106 — Packing k2+1k^2+1 squares in a square

About 51 years old · traced to

Draw nn squares inside the unit square with pairwise disjoint interiors, and let f(n)f(n) be the maximum possible sum of their side lengths. Is f(k2+1)=kf(k^2+1)=k?

References

Additional references

P. Erdős and R. L. Graham, On packing squares with equal squares, Journal of Combinatorial Theory, Series A 19 (1975), 119–123.

Progress summary

Refreshed
Claimed solved

A proposed packing of seventeen squares exceeds the conjectured total, which would disprove the claim from four dimensions onward, but nobody has independently verified it.

Erdős asked whether the largest possible total side length of k2+1k^2+1 squares packed in a unit square is exactly kk, with arbitrary orientations; the case k=4k=4 is f(17)=4f(17)=4.

Known results

  • Erdős proved f(2)=1f(2)=1.
  • Newman proved f(5)=2f(5)=2.
  • Cauchy–Schwarz gives f(k2)=kf(k^2)=k.
  • Halász and Praton obtained reductions and bounds for the broader family f(k2+2c+1)f(k^2+2c+1); Baek, Koizumi, and Ueoro proved the axis-parallel analogue g(k2+1)=kg(k^2+1)=k in 2024.

July 29, 2026 candidate counterexample

A repository gives exact rational coordinates for seventeen squares with total side length 4.000124312403920968>44.000124312403920968>4, claiming f(17)>4f(17)>4 and, via known monotonicity, failure for every k≥4k\ge4. It explicitly says the construction has not been checked by a human referee; AlphaEvolve's separate search did not produce it.

Current status (as of September 2026): The axis-parallel case is proved, while the unrestricted conjecture remains unsettled and the proposed k=4k=4 counterexample is unverified.

Sources

Solutions 0

No solutions have been posted yet.