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Consider the following matrix equation in $n \times n$ circulant $\pm 1$ matrices $A$, $B$, $C$

$$2AA^T+BB^T+CC^T=(4n+4)I-4J$$

where $I$ is the $n \times n$ identity matrix and $J$ is the $n×n$ matrix of ones. My questions:

  1. Are all orders good? Here the term "good" means that the equation has at least one solution.

  2. How to find $A$, $B$ and $C$ when $n$ is given?

Any comment/answer would be appreciated.

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    $\begingroup$ If $n+1$ is the order of a Hadamard matrix, then normalising and removing the row and column of all-ones gives a matrix such that $MM^{\top} = (n+1)I - J$. Taking $A, B, C$ all equal to $M$ gives a solution. For question (2) the techniques used for finding Williamson matrices could probably be adapted to find solutions computationally. $\endgroup$
    – Padraig Ó Catháin
    Commented Dec 29, 2023 at 18:50
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    $\begingroup$ Have you tried writing the $\binom{n+1}{2}$ equations in $3n$ binary unknowns for, say, $n=3$? $\endgroup$
    – Rodrigo de Azevedo
    Commented Dec 29, 2023 at 20:50
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    $\begingroup$ You have $\binom{n+1}{2}$ equations in $3n$ binary unknowns. For $n=3$, you have $6$ equations in $9$ binary unknowns. For $n=4$, you have $10$ equations in $12$ binary unknowns. For $n=5$, you have $15$ equations in $15$ binary unknowns. What about for $n=6$? Do you get the idea? $\endgroup$
    – Rodrigo de Azevedo
    Commented Dec 29, 2023 at 22:51
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    $\begingroup$ For what it worth, in terms of accompanying polynomials of circulants $A, B, C$, the equation reads as $2 a(x)a(1/x)+b(x)b(1/x)+c(x)c(1/x)\equiv 4n+4-4(1+x+...+x^{n-1})\pmod {x^n-1}$, where $a, b, c$ are polynomials of degree $n-1$ with coefficients $\pm 1$. $\endgroup$
    – Fedor Petrov
    Commented Dec 30, 2023 at 10:45
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    $\begingroup$ From the linked question, you can see that if you take $A$, $B$ $C$ to be circulant matrices with maximal-length sequences as their first row this gives an infinite set of solutions for when $n=2^k-1,k\geq 2.$ $\endgroup$
    – kodlu
    Commented Dec 30, 2023 at 15:05

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