If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$?

Question

From the Pre-Regional Mathematics Olympiad, 2019:

If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$?

I have provided one answer below, and would be interested in alternative solutions.

Answer 1

We can rewrite $x =\sqrt{2} + \sqrt{3} + \sqrt6$ as $$\begin{aligned}(x-\sqrt2)^2 &= (\sqrt2 + \sqrt6)^2\\ x^2 - 2\sqrt2x+2 &= 9 + 6\sqrt2\\ x^2-7&=2\sqrt2\space(x+3)\end{aligned}$$ On squaring both sides of the equation, $$x^4-22x^2-48x-23=0$$ Therefore, $a = 0, b = -22, c = -48, d = -23$, implying $|a+b+c+d| = 93$

Answer 2

The only part missing in the existing answer is the fact that the minimal polynomial of $a=\sqrt{2}+\sqrt{3}+\sqrt{6}$ has degree $4$, i.e. the degree of the extension $|\Bbb{Q}[a]:\Bbb{Q}|$ is $4$. Since $a$ is a root of an integer polynomial of degree $4$, the degree of the extension is $1, 2$ or $4$. It cannot be $1$ since $a$ is not rational. If the degree is $2$ then for some rational $p,q$ you have $a^2+pa+q=0$ or $11+2(\sqrt{6}+2\sqrt{3}+3\sqrt{2})+pa+q=0$. But that is impossible. So, indeed, the degree is $4$.

Another way to prove it is to consider homomorphisms from $\Bbb{Q}[a]$ to $\Bbb{Q}[\sqrt{2},\sqrt{3}]$ (which has degree $4$). There are $4$ of those.

Answer 3

Using a bit of Galois theory: first, note that the intermediate fields between $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ are $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, and $\mathbb{Q}(\sqrt{6})$, none of which contains $\sqrt{2} + \sqrt{3} + \sqrt{6}$. Therefore, $\sqrt{2} + \sqrt{3} + \sqrt{6}$ is a primitive generator of the extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ of $\mathbb{Q}$, so the degree of its minimal polynomial is $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] = 4$.

Now, also note that if the minimal polynomial is $p(x) = x^4 + ax^3 + bx^2 + cx + d$, then $a + b + c + d = p(1) - 1$. On the other hand, we also know that $p(x) = \prod_{\sigma \in G} (x - \sigma(\alpha))$ where $G = \operatorname{Gal}(\mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q})$ and $\alpha = \sqrt{2} + \sqrt{3} + \sqrt{6}$. Therefore, $p(1) = N^{\mathbb{Q}(\sqrt{2}, \sqrt{3})}_{\mathbb{Q}}(1 - \alpha)$.

But also, $$ N^{\mathbb{Q}(\sqrt{2}, \sqrt{3})}_{\mathbb{Q}} (1 - \alpha) = N^{\mathbb{Q}(\sqrt{2})}_{\mathbb{Q}} N^{\mathbb{Q}(\sqrt{2}, \sqrt{3})}_{\mathbb{Q}(\sqrt{2})} [(1 - \sqrt{2}) - (1 + \sqrt{2}) \sqrt{3}] = \\ N^{\mathbb{Q}(\sqrt{2})}_{\mathbb{Q}} [(1 - \sqrt{2})^2 - 3 (1 + \sqrt{2})^2] = N^{\mathbb{Q}(\sqrt{2})}_{\mathbb{Q}} (-6 - 8 \sqrt{2}) = 6^2 - 2 \cdot 8^2 = -92. $$ Thus, $a + b + c + d = -93$, and $|a + b + c + d| = 93$.