Prove that in a field $F, \exists a,b,c \in F$ such that $x^3+x^2+1$ is a divisor of $x^{2018} + ax^3 + bx + c$

Question

I have no idea how to even start this! Maybe using the Euclidean algorithm and showing that the extra term is 0?

Answer 1

Apply Euclidean algorithm to $x^{2018}$ and $x^3+x^2+1$, we can find a polynomial $q(x)$ and quadratic polynomial $\alpha x^2 + \beta x + \gamma$ such that

$$x^{2018} = q(x)(x^3+x^2+1) + \alpha x^2 + \beta x + \gamma\\ \Downarrow\\ x^{2018} + ax^3 + bx + c = q(x)(x^3+x^2+1) + \underbrace{ax^3 + \alpha x^2 + (\beta+b)x + (\gamma + c)}_{\mathcal{M}}$$ If we take $a = \alpha, b = -\beta, c =\alpha - \gamma$, we have $$\mathcal{M} = \alpha (x^3 + x^2 + 1) \quad\implies\quad x^{2018} + ax^3 + bx + c = (q(x)+\alpha)(x^3+x^2+1)$$ For such a choice of $a,b$ and $c$, $x^3+x^2+1$ is a factor of $x^{2018} + ax^3 + bx + c$.