---

Setting

$F_k(N)$ and the constants $c_k^-\le c_k^+$ are as on the Theorem 1.2 page (pp. 1 and 3), and $c=0.171500\ldots$ is the Hall--Montgomery constant (1.2) (p. 2).

Statement

Remark 1.3 (pp. 3--4). The paper says its method would give an explicit lower bound for $c_k^-$, but a small one, much smaller than $c$ and deteriorating as $k\to\infty$. It then records an argument communicated by Csaba Sándor:

(a) for all $k,l\ge1$,

$$F_{k+l}(N)\le\max\bigl(F_k(N),\,F_l(N)+k\bigr),$$

since a set larger than the right side has $k$ distinct elements with square product, and after their removal still has $l$ more (p. 3);

(b) dividing by $N$ and letting $N\to\infty$, for all $k,l\ge1$ and either choice of sign,

$$c_{k+l}^{\pm}\ge\min\bigl(c_k^{\pm},\,c_l^-\bigr)$$

(p. 4);

(c) for odd $k$, $c_k^{\pm}\le c<c_2^-=1-6/\pi^2$, so (b) with $l=2$ gives $c_{k+2}^{\pm}\ge c_k^{\pm}$: $c_k^{\pm}$ is nondecreasing along odd $k$ (p. 4). The print writes the value of $c_2^-$ in this sentence as $1-\frac{\pi^2}{6}$; the value it gives on p. 3 is $c_2^-=c_2^+=1-\frac6{\pi^2}=0.39207\ldots$, which is the one the comparison needs.

The remark ends with conjectures, not results (p. 4): it calls it plausible that $c_k^-=c_k^+$ tends to $c$ as $k\to\infty$ through odd values, and more boldly that $c_k^-=c_k^+=c$ for every odd $k\ge5$; by the monotonicity, the bolder conjecture would follow from the case $k=5$ if $F_k(N)/N$ has a limit. It says the limited numerics of Figure 1 (p. 4) are not inconsistent with this.

Proof pointer

The argument for (a) is the two-step removal sketched in the remark itself (pp. 3--4); (b) and (c) follow by taking limits and comparing constants.

Read depth

Claims checked: the remark was read clause by clause on the page images of pp. 3--4 of the print. Nothing here is independently reviewed.

Source. Terence Tao, On product representations of squares, Acta Math. Hungar. 175 (2025), no. 1, 142--157, doi:10.1007/s10474-025-01505-7; preprint arXiv:2405.11610. Labels and pages are those of arXiv:2405.11610v3, the edition named on the source card.

Bears on

• Problem 121: the problem asks whether $F_{2k+1}(N)=(1-o(1))N$. The remark's monotonicity says that for odd sizes the proportion missed, $c_k^{\pm}$, does not decrease as $k$ grows by $2$; it gives no positive lower bound by itself, which comes from Theorem 1.2. The conjecture $c_k^-=c_k^+=c$ for odd $k\ge5$ is stated in the paper without proof.