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Claim. The answer to Problem 106 is no: $f(17)>4$, so $f(k^2+1)=k$ fails at $k=4$. The witness is an explicit packing of seventeen squares in the unit square, sixteen of them parallel to its sides and one tilted by about $7$ degrees, with side lengths summing to

$$\frac{2190452873}{547596200}=4.000124312\ldots$$

Two parameters determine the whole layout; they solve two linear equations saying that the tilted square touches a corner of each of two neighbors. The tilt supplies the entire excess: with the tilted square replaced by one parallel to the sides, the sum is exactly $4$, in line with the theorem of Baek, Koizumi and Ueoro that packings of $k^2+1$ squares parallel to the sides reach exactly $k$. Containment of all seventeen squares in the unit square and disjointness of the interiors of all $136$ pairs were checked in exact rational arithmetic. Since $k\bigl(f(k^2+1)-k\bigr)$ is nondecreasing in $k$, as Raj Singh observed, one counterexample at $k=4$ gives a constant $c>0$ with $f(k^2+1)\geq k+c/k$ for every $k\geq4$.

Submission note. Posted to erdosproblems.com as a proof claim by Conner Silverstein (account Sprite144) on 29 July 2026, giving "Claude Opus 5 (construction, search, exact verification); ChatGPT and Grok (independent review)" as the AI used:

I claim f(17) > 4, so f(k^2+1) = k is false at k = 4. An explicit arrangement of 17 squares in the unit square: sixteen aligned with the edges, one tilted about 7 degrees. Side lengths total 2190452873/547596200 = 4.000124312..., which beats 4. It comes from a formula, not a search: two numbers fix the whole layout, found by solving two linear equations saying the tilted square presses against two neighbours' corners. The tilt is the whole margin. Straight, that square reaches only 0.173122157 in its gap; tilted, 0.173246470. Swap the straight one back in and the total is exactly 4, the known answer when every square is aligned. Checked in exact fractions, no decimals: all 17 inside the unit square, none of the 136 pairs overlap. Not refereed. Notes: Prior computational work: AlphaEvolve searched this problem with the angle as a free parameter (arXiv:2511.02864, section 35) and matched only 4 at n = 17. A likely explanation is that this configuration has 36 exact tangencies and squares touching the walls with clearance exactly 0, so it sits precisely on the valid/invalid boundary of a floating-point intersection test that scores invalid configurations as minus infinity. The formulation used here fixes the angles, which makes the problem a linear program, so tangencies appear as active constraints rather than near-violations. Verification code (exact rational arithmetic, standard library only) is in the linked gist: construct.py rebuilds the configuration from the 2x2 system, verify_independent.py checks it by convex polygon clipping, witness.py emits the 136 separating-axis witnesses.

Claimant. Conner Silverstein, posting under the forum account Sprite144, submitted the claim on 2026-07-29 with the repository linked above. The claim names Claude Opus 5 for the construction, the search and the exact verification, and ChatGPT and Grok for independent reviews; the site credits the result to Claude Opus 5 prompted by Silverstein.

Acceptance. Thomas Bloom, the site's curator, marks the problem disproved and credits this result on the problem page (edited 2026-08-28). The site's label notes a Lean verification, but neither the claim nor the problem page links one; the Lean proof of $f(17)>4$ in Boris Alexeev's repository, credited to Raj Singh, uses a different packing and is recorded on its own claim page. This claim is accepted on the curator's documented acceptance alone. No refereed publication exists. The problem's discussion also reports a simpler construction by Bojan Bašić with $k=16$, posted as a comment only.

What remains. $f(5)=2$ is known; whether $f(10)=3$ is open, and the general conjecture of Erdős and Soifer and of Campbell and Staton, which Praton showed equivalent to $f(k^2+1)=k$ for all $k$, fails with it.