Is there a 12x12 symmetric Hadamard matrix?

Question

More generally, is there a simple condition for which $n$ there are symmetric Hadamard matrices of order $n$?

This set of $n$ is closed under multiplication via the Kronecker product.

Answer 1

There is a construction that generates a symmetric Hadamard matrix of order $2(q+1)$ for any prime $q\equiv1\bmod4$ (see Wikipedia, especially Paley Construction II). Since $q=5$ qualifies, the required order-12 matrix is a solved problem. The construction may also be modified for the case where $q$ is one greater than a multiple of $4$ and a power of a prime, like $q=3^2=9$ giving a symmetric Hadamard matrix of order 20.

Answer 2

It seems that the conjectures for (skew-)symmetric Hadamard matrices are "the same" : for every $n\equiv 0 \mod 4$ there exists a (skew-)symmetric Hadamard matrix of size $n$ ( skew-symmetric means $H+H^t = 2 I$). Check out this talk, especially p.20. Good luck!

$\bf{Added:}$ It would be interesting to see how the Paley I $H_1$ and Paley II $H_2$ constructions are related in the case $n=12$. The first is skew-symmetric, the second symmetric. Sloane says they are equivalent. Maybe there exist $D$ a $\pm 1$ diagonal matrix, $P$ a permutation matrix, such that

$$H_2 = H_1 \cdot D \cdot P$$

Another case where both Paley I and Paley II work is for $n=60 = 59 + 1 = 2(29+1)$.