We want to generate a series of invertible, symmetric integer matrices such that the sum of the odd powers of its eigenvalues are zero. Is there a nice, well known set of matrices that satisfy this, or is there a way to generate such matrices?
We have been looking at some examples and this is the only interesting condition we found, and would like to look at a few more examples to see what is going on.
Call your matrix $A$. For each positive real number $\lambda>0$, let $m(\lambda)$ be the difference between the multiplicities of $\lambda$ and $-\lambda$ in the spectrum of $A$, where the multiplicity of a non-eigenvalue is understood to be zero. Let $\{\lambda_1,\lambda_2,\ldots,\lambda_k\}$ be complete set of distinct positive numbers such that $m_i:=m(\lambda_i)\ne0$. Then $$ M=\pmatrix{m_1\lambda_1&m_2\lambda_2&\cdots&m_k\lambda_k\\ m_1\lambda_1^3&m_2\lambda_2^3&\cdots&m_k\lambda_k^3\\ \vdots&\vdots&&\vdots\\ m_k\lambda_1^{2k-1}&m_2\lambda_2^{2k-1}&\cdots&m_k\lambda_k^{2k-1}} =VD $$ where $V$ is the Vandermonde matrix generated by $\lambda_1^2,\lambda_2^2,\ldots,\lambda_k^2$ and $D=\operatorname{diag}(m_1\lambda_1,\ldots,m_k\lambda_k)$. Hence $M$ must be empty, or else it is singular (because the all-one vector still lies inside its kernel) but $V$ and $D$ are not. Thus $m(\lambda)$ is identically zero, i.e., for each eigenvalue $\lambda$ of $A$, $-\lambda$ is also an eigenvalue of the same multiplicity.
Hence $A$ must be even-sized and its eigenvalues occur in pairs of the form $(\lambda,-\lambda)$. It follows that $A$ can be decomposed over $\mathbb R$ as $$ A=Q\left[\pmatrix{a_1&b_1\\ b_1&-a_1}\oplus\pmatrix{a_2&b_2\\ b_2&-a_2}\oplus\cdots\oplus\pmatrix{a_n&b_n\\ b_n&-a_n}\right]Q^T\tag{$\ast$} $$ where $Q$ is orthogonal and $\pmatrix{a_i&b_i\\ b_i&-a_i}\ne0$ for each $i$. So, one way to generate a desired matrix is as follows: