If the degree of $\alpha=5555$, what is the degree of $\alpha^2$?

Question

$\alpha$ is an algebraic number with degree $5555$. What is the degree of $\alpha^2$?

Here are my thoughts so far:

I think it is true that $\mathbb{Q} \subseteq \mathbb{Q}[\alpha^2] \subseteq \mathbb{Q}[\alpha]$.

There is a theorem which says that $|\mathbb{Q}[\alpha]:\mathbb{Q}|=\deg(\alpha)$ and the Tower Rule tells us that $|\mathbb{Q}[\alpha]:\mathbb{Q}|=|\mathbb{Q}[\alpha]:\mathbb{Q}[\alpha^2]||\mathbb{Q}[\alpha^2]:\mathbb{Q}|=\deg(\alpha)=5555$. So we know that $\deg(\alpha^2)$ divides $5555$.

However, I don't know how to narrow down the options. $\deg(\alpha^2)$ could equal any of $1, 5, 11, 55, 101, 505, 1111,5555$ though I'm fairly certain it's not going to be $1$.

Is there a way to deduce $\deg(\alpha^2)$ based on just what we know here or do we need some other knowledge too?

Thanks in advance.

Answer 1

The number $5555$ is a red herring. You just need the fact it's odd.

Since $\def\Q{\mathbb{Q}}\Q(\alpha^2)\subseteq\Q(\alpha)$, you have $$ [\Q(\alpha):\Q]=[\Q(\alpha):\Q(\alpha^2)]\,[\Q(\alpha^2):\Q]. $$ What can the degree of $\alpha$ over $\Q(\alpha^2)$ be? Consider the polynomial $X^2-\alpha^2$ which has coefficients in $\Q(\alpha^2)$, in order to give an answer.

Since the polynomial has degree $2$, the degree of $\alpha$ over $\Q(\alpha^2)$ is either $1$ (if the polynomial is reducible) or $2$, because certainly $\alpha$ is a root of $X^2-\alpha^2$.

If $[\Q(\alpha):\Q]$ is odd, what more can you say?