$\det(A^4+I)=29$ is not solvable by any $A\in M_4(\mathbb Z)$ I recently encountered the following problem.

Given any $A \in M_4(\mathbb Z)$, show that $\det(A^4+I)\ne29$, where $I$ denotes the identity matrix.

LHS can be written as the product of $1+{\lambda _i}^4$ where $\lambda _i$ denotes the eigenvalues of A. By using AM-GM inequality, I found that A is either invertible in $M_4(\mathbb Z)$ or has a zero determinant. I cannot go further. Can anyone help me?

Consider the matrix $A$ modulo $29$.
The matrix $A^4+I$ has determinant $29$ so has rank $3$ and nullity
zero considered as a matrix over $\Bbb F_{29}$. It has a unique null-vector
$u$ up to scalar multiplication: $(A^4+I)u\equiv 0\pmod{29}$
and $(A^4+I)v\equiv 0\pmod{29}$ implies $\newcommand{\la}{\lambda}v\equiv\la u\pmod{29}$.
In particular, taking $v=Au$, we find $Au=\la u\pmod{29}$ for some
$\la$. Then $\la^4+1\equiv0\pmod{29}$. But as $8\nmid(29-1)$, this congruence is insoluble.
