The odd powers of the adjacency matrix have all 0's on the diagonal. So this means the sum of the $i$-th powers of the eigenvalues is 0 for each odd $i$. Since the adjacency matrix is symmetric, it has real eigenvalues. So the eigenvalues are real numbers $\lambda_1,\dots ,\lambda_n $ with $\sum_{j=1}^n \lambda_j^i = 0$ for each odd $i$.
I am guessing this probably should mean that the nonzero eigenvalues come in pairs of equal magnitude and opposite sign. I wonder if there is a good trick for efficiently proving this sort of thing -- about collections of real numbers satisfying such an infinite family of relations? (If so, I don't know this trick.)