3

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

John Smith
  • 457
  • Yes, just try this way... More precisely, adjust $a$, $b$ and $c$ to obtain a rest equal to $0$. – J.-E. Pin Apr 22 '18 at 16:59
  • @J.-E.Pin Could you please explain a little more. I have that by the Euclidean algorithm, there exist $f,r \in F[x]$ such that $x^{2018} + ax^3 + bx + c$ = $(x^3+x^2+1)f(x) + r(x)$, where the degree of $r(x)$ is less than 2018, but I don't get how to manipulate $a,b,c$ – John Smith Apr 22 '18 at 17:09
  • The degree of $r(x)$ is $< 3$. – J.-E. Pin Apr 22 '18 at 17:23
  • Oh right. Thank you, but I still don't see how to manipulate $a,b,c$? I know that $f = x^{2015} + ...$ so we need something to get rid of the $x^{2016}, x^{2015}$ terms, but I don't see how – John Smith Apr 22 '18 at 17:29

1 Answers1

3

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 $x^{2018} + ax^3 + bx + c$.

achille hui
  • 122,701
  • 1
    This is perfect thank you! It makes perfect sense, but how did you know that you should just take the $x^{2018}$ term for the Euclidean algorithm? – John Smith Apr 22 '18 at 17:33
  • @JohnSmith Instead of $x^{2018}$, you can use any $x^{2018} + a_0 x^3 + b_0x + c_0$ for the Euclidean algorithm. Since there isn't other obvious choice one should use, I just pick the simplest one. – achille hui Apr 22 '18 at 17:40