Centered four-cycle identity for a biregular graph ($t_4(G)={\delta }^4+n^{-4}\mathrm{tr}((H{H}^{\mathsf{T}}{)}^2)$)

For an equal-part biregular graph with part size $n$ and degree $\delta n$, let $M$ be its biadjacency matrix and $H=M-\delta J$, where $J$ is the all-ones matrix. The constant row and column sums make $H$ annihilate constants on both sides. Since $M{M}^{\mathsf{T}}={\delta }^{2}nJ+H{H}^{\mathsf{T}}$ and the two summands have zero products, their squared matrix trace splits as displayed. The bipartite four-cycle density counts the squared matrix trace of $M{M}^{\mathsf{T}}$ divided by $n^4$.