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I'm interested in finding real algebraic numbers $a$ of large absolute value satisfying the property that $\mathbb{Q}(a) = \mathbb{Q}(a^2)$. A more precise question is: for which positive integers $d$ is it possible to find real algebraic numbers $a$ of degree $d$ over $\mathbb{Q}$ of arbitrarily large absolute value such that $\mathbb{Q}(a) = \mathbb{Q}(a^2)$? For my purposes, ideally I would like to know that this is possible for every $d$.

Since $[\mathbb{Q}(a):\mathbb{Q}(a^2)]\le 2$, this is trivially possible if $d$ is odd (take any element of degree $d$ and consider the elements $a+N$ for $N\in\mathbb{Z}$). The situation for even $d$ seems to be more tricky.

stupid_question_bot
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    In the post here shows that $x^n-x-1$ is irreducible over $\mathbb{Q}$ for all $n\geq 2$. If $n=2k$, it follows that if $a$ is a root, then $a=(a^2)^k-1$, so $\mathbb{Q}(a)=\mathbb{Q}(a^2)$. Then $a+N$ is a root of $(x-N)^{2k}-(x-N)-1$, and the same argument applies. – Arturo Magidin May 20 '23 at 06:25
  • Hmmm... you may need to be more careful. E.g., if $a=\sqrt{2}+N$, $N\neq0$, then $a^2=2N\sqrt{2} +(N^2+2)$, so $\mathbb{Q}(a)=\mathbb{Q}(a^2)$. But $b=a-N=\sqrt{2}$ has $\mathbb{Q}(b^2)\neq\mathbb{Q}(b)$. – Arturo Magidin May 20 '23 at 14:24
  • Also why must the root be real? – stupid_question_bot May 20 '23 at 14:46
  • The value at $0$ is negative and the value at $2$ is positive, so it has a positive real root for sure. – Arturo Magidin May 20 '23 at 15:17

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Not only is it true for any $d$, it is true for any number field $K$ which has at least one real embedding (so can be viewed as a subfield of $\mathbf{R}$). Note that there exists such a field for each $d$, e.g. $K = \mathbf{Q}(2^{1/d})$.

By the primive element theorem, we can write $K = \mathbf{Q}(b)$ viewed as a subfield of $\mathbf{R}$. Then I claim that

$$K = \mathbf{Q}((b + r)^2)$$

for all but finitely many rational numbers $r$. In particular taking $r$ to be a very big rational number and $a = (b + r)$, then

$$K = \mathbf{Q}(a) = \mathbf{Q}(a^2)$$

and $a$ can be as large as you want.

The proof is just the same as the proof of the primitive element theorem. Namely let $K_r = \mathbf{Q}((b + r)^2)$. We may assume there are infinitely many $r$ such that $K_r \ne K$. Since $K/\mathbf{Q}$ has only finitely many subfields (the key point), it follows that two of these fields coincide so $K_r = K_s$ are both proper subfields of $K$. But if $K_r = K_s$ then these fields contain $(b+r)^2$ and $(b+s)^2$, and so also

$$\frac{(b + r)^2 - (b + s)^2}{2(r - s)} - \frac{(r + s)}{2} = b,$$

but then they contain $b$ and so contain $K$, a contradiction.

user297024
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    For the benefit of aged mathematicians with failing vision, never use “$a$” and “$\alpha$” in close proximity to each other. (But I like your proof) – Lubin May 20 '23 at 18:57