Discriminant of characteristic polynomial as sum of squares

The answer for a general $n$ is positive: the discriminant is a sum of squares of polynomials in the entries of $H$. The first formula was given by Ilyushechkin and involves $n!$ squares. This number was improved by Domokos into $$\binom{2n-1}{n-1}-\binom{2n-3}{n-1}.$$ See Exercise #113 on my page.

Details of Ilyushechkin's solution. Consider the scalar product $\langle A,B\rangle={\rm Tr}(AB)$ over ${\bf Sym}_n({\mathbb R})$. It extends as a scalar product over the exterior algebra. Then the discriminant equals $$\|I_n\wedge H\wedge\cdots\wedge H^{n-1}\|^2,$$ which is a sum of squares of polynomials.

The answer is Yes in any dimension by a result of Ilyushechkin in Mat. Zametki, 51, 16-23, 1992.

See my previous MO answer

real symmetric matrix has real eigenvalues - elementary proof

We know that $H$ is symmetric, and therefore, diagonalizable, as $H = Q^TDQ$ for some orthogonal matrix $Q$. Moreover, $D$ and $Q$ have the same eigenvalues, and thus the same characteristic polynomials. Perhaps this can be used?

In any case, this reference by Domokos mentions the other answers and references as well. It gives some explicit expressions in the 3x3 case, both in five squares (theorem 7.3) and in seven squares (theorem 7.4), showing that the decomposition is not unique.