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user369335
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Currently, this is an open problem. So I only got a partial answer.

We just need to consider the following two different situations:

Case 1: $n$ is not a perfect square. There is no solution if $\lvert trace(H) \rvert \neq 0$; and $0$ is a possible choice for $\lvert trace(H) \rvert$ if and only if symmetric HM conjecture is correct.

Case 2: $n$ is a perfect square. There is no solution if $\lvert trace(H) \rvert \notin \{0,2\sqrt{n},4\sqrt{n},...,n\}$; and all possible choices are known only for small cases, for example:

$n$ All possible choices for $\lvert trace(H) \rvert$
$1$ $ 1 $
$4$ $ 0,4 $
$16$ $ 0,8,16 $
$36$ $ 0,12,24,36 $
$64$ $ 0,16,32,48,64 $
user369335
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