Find all polynomial $f(x)$ such that $x^3-1\mid f(x)g(x)-1$ This is a problem that has haunted me for more than some days. Not all the time - but from time to time, and always on windy or rainy days, it suddenly reappears in my mind:
Find all  polynomial $f(x)$,such $f(x)\in Z[x],\deg{(f(x))}\le 2$,and there exist $g(x)\in Z[x]$,such
$$x^3-1\mid f(x)g(x)-1$$
I have $f(x)=x^2,x$ ,have other solution,and How to find all solution?. Thanks

Let $\rho = e^{2\pi i/3} = \frac{-1 + i\sqrt{3}}{2}$. In $\mathbb{C}[x]$, we have the factorisation $x^3-1 = (x-1)(x-\rho)(x-\rho^2)$, so if $x^3-1 \mid f(x)g(x) - 1$, we must have $f(1)g(1) = f(\rho)g(\rho) = f(\rho^2)g(\rho^2) = 1$. Since $f,g$ are assumed to be polynomials with integer coefficients, we know that we must have $f(1) = g(1) = \pm 1$, and the values at $\rho^k$ belong to the ring $R = \mathbb{Z}[\rho]$ of Eisenstein integers, so $f(\rho^k)$ must be a unit in $R$. From the ansatz $f(x) = ax^2 + bx + c$ we obtain the conditions
\begin{gather}
a + b + c = \pm 1, \\
a\rho^2 + b\rho + c \in R^{\times}, \\
a\rho^4 + b \rho^2 + c \in R^{\times}.
\end{gather}
With $\rho^2 = -(1+\rho)$ and $\rho^4 = \rho$, that becomes
\begin{gather}
a+b+c = \pm 1, \\
(c-a) + (b-a)\rho \in R^{\times}, \\
(c-b) + (a-b)\rho \in R^{\times}.
\end{gather}
The units of $R$ are $1,\, 1 + \rho,\, \rho,\, -1,\, -1-\rho,\, -\rho$, so we must have either $a = b$ or $a - b = \pm 1$. Since $(-f(x))(-g(x)) = f(x)g(x)$, we can ignore the case $a - b = -1$.
In the case $a = b$, we must have $c-a = \pm 1$, and thus the first condition yields $a + b + c = 3a \pm 1 = \pm 1$, so it follows that $a = 0$, and we have the trivial option $f(x) = \pm 1$, with $g(x) = \pm x^3$ as a possible companion.
In the case $a - b = 1$, it follows that $c-a = -1$ or $c-a = 0$ from the second condition, and the corresponding $c - b = c - a + 1 = 0$ or $c - b = c - a + 1 = 1$ then also satisfies the third condition. So we either have $c = b$ or $c = a$ then. In the first condition, $c = b = a-1$ leads to $3a - 2 = \pm 1$, thus $a = 1$ and $b = c = 0$, which yields $f(x) = x^2$. The possibility $c = a$ leads to $3a - 1 = \pm 1$, hence $a = 0$, and therefore $f(x) = -x$.
Picking up the negatives that we dropped by ignoring $a - b = -1$, the complete list of $f \in \mathbb{Z}[x]$ with $\deg f \leqslant 2$ such that there is a $g \in \mathbb{Z}[x]$ with $x^3 - 1 \mid f(x)g(x) - 1$ is
$$f(x) = \pm 1 \quad\text{or}\quad f(x) = \pm x \quad\text{or}\quad f(x) = \pm x^2.$$
