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.$$ Now what about the eigenvalues of such a matrix? In the case of binomial coefficients, they turned out to be integers in the cited question, and more precisely $2^1, 2^2, \ldots, 2^n$. …

Can such a graph be integral, i.e. have only integer eigenvalues? … But among the eigenvalues... the first prime ($-19$) only appears at $d=133$ and is so far the only one! So far, all prime factors of the eigenvalues divide at least one of $a,b,c,d$. …

Denote by $c(M)$ the number of pairs of non-real eigenvalues in $TS(M)$. By definition, $c(M)\leqslant n! \cdot [n/2]$. For given $n$, what is $\max c(M)$? … For $n=4$, we have $c(M)\leqslant48$, but it seems like $\max c(M)=36$, which is made plausible not only by the graphic below but also by the observation that the eigenvalues of the extremal matrices seem …

From Cardano's formula we get as a necessary condition for integer eigenvalues of this $3\times3$ matrix that $$\Delta=(ab+ac+bc)^3-27(abc)^2$$must be a square. … $K_{8,15,32,40\ \ }$ or $K_{3,35,48,77\ \ }$, which have two integer nonzero eigenvalues. Do such 4-partite integral graphs exist? …

\cdot n$ eigenvalues be real? Denote by $c(M)$ the number of pairs of non-real eigenvalues in $TS(M)$. For a matrix of rank 1, its TS is trivially real. … $J=J_n$ denotes the all-1-matrix and $I=I_n$ the unit matrix, it is easy to show that $c(J+\epsilon I)=c(I)$ for all $\epsilon\in\mathbb R$ (corresponding permutations of both matrices have the same eigenvalues …