$\det(A^4+I)=29$ 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?

This answer is a one I was taught from my friend.
$\det(A^4+I)=\det(A^2+I+\sqrt{2}A)\det(A^2+I-\sqrt{2}A)$.
We can put $\det(A^2+I+\sqrt{2}A)=a+b\sqrt{2}$ and then $\det(A^2+I-\sqrt{2}A)=a-b\sqrt{2}$ where $a,b\in\mathbb{Z}$. Hence we obtain $a^2-2b^2=29$, which has no solution.
