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I am currently reading "On Subsets with Cardinalities of Intersections Divisible by a Fixed Integer" by P. Frankl And A. M. Odlyzko. They used the following result without citation:

For each number $l$, we have a decomposition $l=l_1+\cdots+l_q$ with $l_i\geq\epsilon l$ for a fixed constant $\epsilon$, such that the Hadamard matrix of size $4l_k$ exists.

Would anyone provide a reference or a short proof that I missed? THX

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  • $\begingroup$ I thought Hadamard matrices existed for each size $4n$? $\endgroup$ Commented May 4, 2023 at 20:32
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    $\begingroup$ @CarloBeenakker, that is the Hadamard Conjecture, widely believed but open $\endgroup$
    – kodlu
    Commented May 4, 2023 at 21:07
  • $\begingroup$ The paper actually says $l$ is even, you should transcribe more carefully $\endgroup$
    – kodlu
    Commented May 4, 2023 at 21:14
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    $\begingroup$ Is there supposed to be a "for all $k$, $1\le k\le q$"? $\endgroup$ Commented May 5, 2023 at 3:41
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    $\begingroup$ It should follow from some version of Goldbach (since there exist Hadamard matrices of order $2(p+1)$ for all odd prime $p$, this follows from Goldbach with many summands but additional condition that they are all of comparable size.) $\endgroup$ Commented May 5, 2023 at 6:04

1 Answer 1

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This can be proved using the strategy of Fedor Petrov and a theorem from the following paper:

Haselgrove, C. B., Some theorems in the analytic theory of numbers, J. Lond. Math. Soc. 26, 273-277 (1951). ZBL0043.04704.

Let $63/64 < \theta < 1$. According to Theorem A of Hasselgrove, if $m$ is a sufficiently large odd number, then $m$ is the sum of three primes $p_1$, $p_2$, $p_3$ with $|p_i-m/3| < m^{\theta}$. Putting $m = 2 \ell-3$, we obtain $$4 \ell = 2(p_1+1) + 2(p_2+1) + 2(p_3+1).$$

The Paley construction gives a Hadamard matrix of size $2(p+1)$ for any odd prime $p$. Since $p_i = (1/3) m + O(m^{\theta})$, we have $2(p_i+1) = (4/3) \ell + O(\ell^{\theta})$.

The exponent $63/64$ has been improved on by many other authors; for example, Matomakai, Maynard and Shao push it down to $11/20$. But the OP just asked for each of the matrix sizes to be comparable to $\ell$, so I'll stop here.

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  • $\begingroup$ A few weeks ago I was at the talk of Mark Rudelson, and he and Xiaoyu Dong 1 actually used this same idea to construct almost Hadamar matrices of all sizes (including the case $n$ not divisible by $4$). $\endgroup$ Commented May 16, 2023 at 8:35
  • $\begingroup$ @AlekseiKulikov Thanks! The relevant work seems to be Section 2 of arxiv.org/abs/2207.07523 . $\endgroup$ Commented May 16, 2023 at 13:16
  • $\begingroup$ Ah, and Dong and Rudelson's result is nicer, in that they get almost Hadamard matrices with entries in $\{ -1, 1 \}$, rather than $\{ -1, 0, 1 \}$. $\endgroup$ Commented May 16, 2023 at 13:20

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