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



About a result on Hadamard matrix
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**](https://doi.org/10.1112/jlms/s1-26.4.273), J. Lond. Math. Soc. 26, 273-277 (1951). [ZBL0043.04704](https://zbmath.org/?q=an:0043.04704).


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](https://en.wikipedia.org/wiki/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](https://arxiv.org/abs/1610.02017) 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.


