Find a matrix $X \in M_{2}(\mathbb Z)$ such that $$X+X^2+X^3=\begin{bmatrix} 1&2005\\ 2006&1\end{bmatrix}$$
My try:
Let $$X=\begin{bmatrix} a&b\\ c&d \end{bmatrix}$$ where $a,b,c,d\in Z$ then $$X^2=\begin{bmatrix} a^2+bc&ab+bd\\ ac+cd&bc+d^2 \end{bmatrix}$$ then $$X^3=\begin{bmatrix} a^3+abc+abc+bcd&a^2b+b^2c+abd+bd^2\\ a^2c+acd+bc^2+cd^2&abc+bcd+bcd+d^3 \end{bmatrix}$$ so $$X+X^2+X^3=\begin{bmatrix} a^3+2abc+bcd+a^2+bc+a&a^2b+b^2c+abd+bd^2+ab+bd+b\\ a^2c+acd+bc^2+cd^2+ac+cd+c&abc+2bcd+d^3+bc+d^2+d \end{bmatrix}$$ then we have $$\begin{cases} a^3+2abc+bcd+a^2+bc+a=1\\ a^2b+b^2c+abd+bd^2+ab+bd+b=2005\\ a^2c+acd+bc^2+cd^2+ac+cd+c=2006\\ abc+2bcd+d^3+bc+d^2+d=1 \end{cases}$$ Note $2005=5\cdot 401 $ is prime number,so
$$b(a^2+bc+ad+d^2+a+d+1)=2005$$ we can $b=\pm 1$ ,or $b=\pm 2005$ ,or $b=\pm 5$ or $b=\pm 401$
and note $2006=2\times 1003$
then $c=\pm 2$ ,or $c=\pm 1003$or $c=\pm 1$,or $c=\pm 2006$
so following is very ugly. But I can't.Thank you ,maybe this problem have other nice methods,Thank you
There's no such $X$, even with rational entries. If there were, then it would have an eigenvalue that's either rational or a quadratic irrationality. But if $\lambda$ is an eigenvalue of $X$ then $\lambda + \lambda^2 + \lambda^3$ is an eigenvalue of $\left[\begin{array}{cc}1&2005\cr2006&1\end{array}\right]$. But those eigenvalues are the roots $x = 1 \pm \sqrt{2005\cdot 2006}$ of $(x-1)^2 = 2005 \cdot 2006$, and the polynomial $(\lambda^3+\lambda^2+\lambda-1)^2 - 2005 \cdot 2006$ turns out to be irreducible, so none of its roots can be the eigenvalue of a $2 \times 2$ matrix with rational entries, QED.