Example of matrices satisfying certain matrix equation

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EDIT : What follows was written as an answer to the initial redaction of the question, where $B$ was supposed to be nilpotent (i. e. $B^k-B=\mathbf{0}$ for some prime $k$, with no specification of $k$). Furthermore, it uses an wrong definition of nilpotency (first-class thinko...). Therefore, it answers a different problem, with very weak relation to the real question. Kept for what it's worth...

Not an answer, but a contribution. Furthermore, I won't try to solve the (hard) problem of solving in $\mathbb{Z}$, but, at least as a first step, to solve it in $\mathbb{Q}$.

The problem has two parameters :

• The matrices' dimension $n$
• the index (degree) of $B$'s nilpotency.

Each of the matrix equations translates to $n^2$ equations involving $2n^2$ unknown equations ; similarly, the nilpotency of $B$ translates to $n^2$ equations of maximal dregree $k$ involving $n^2$ parameters.

However, these equations are far from being independent. For example, the nilpotency of $B$ translates to

• $n=2\, k=2$ : 4 equations of 4 unknowns ; the resulting ideal has dimension 2, which means that this system has two degrees of freedom.

• $n=3\\,k=2$ ; 9 equations of 9 unknowns, the resulting system still has 4 degrees of freedom.

The "obvious" brute-force way (a polynomial system in $a_{i,j}\,b_{i,j}$) doesn't work. However, one may try to solve for $B$ given $A$, i. e. build a system of polynomials in $b_{i,j}$ whose coefficients are taken in (the fraction field of) the ring of polynomials in $a_{i,j}$. For example, after running :

NN=2
k=2
R1=PolynomialRing(QQ, ["a%d%d"%(u, v) for v in range(NN) for u in range(NN)])
# R1.inject_variables()
A=matrix(NN,R1.gens())
R2=PolynomialRing(FractionField(R1), ["b%d%d"%(u, v) for v in range(NN) for u in range(NN)])
# R2.inject_variables()
B=matrix(NN,R2.gens())
S1=(A^2*B+A*B*A+B*A^2).list()
S2=(B^2*A+B*A*B+A*B^2).list()
# Nilpotency
S3=(B^k-B).list()
J1=R2.ideal(S1+S2+S3)

one gets :

sage: R2.ideal(S1+S2+S3).variety()
[{b11: 0, b01: 0, b10: 0, b00: 0}]
sage: J1.dimension()
0
sage: J1.variety()
[{b11: 0, b01: 0, b10: 0, b00: 0}]

This result is easily repeated for $k\in\,(2,3,5,7)$ ; for $k=11$, the system failed to compute J1.dimension() after about 10 minutes (I threw the towel...).

Similarly, in the case $n=3,\,k=2$, the system has been searching the same J1.dimension() for about 3 hours.

FWIW, the (gratis-but-not-free) Wolfram engine's Reduce exhibits a similar behaviour., However, Solve seems to be able to get the (unique) solution (null $B$) for $n=3,\,k=3$ in a reasonable time ; getting the (same) answer for $n=4,\,k=2$ needs about a minute. Going beyond $n=4$ or $k=7$ leads to unreasonable (=exceeding my patience) times.

This brute-force symbolic approach using only the polynomials machinery (or whatever the Wolfram engine uses) seems bound to fail. An approach using more results of linear algebra is needed.

HTH,