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Do there exist rational numbers $a,b\in\Bbb{Q}$ such that $\sqrt[4]{2}=a+b\sqrt{1+\sqrt{2}}$?

I am pretty sure that there do not but I cannot figure out how to show such. I have tried manipulating this equation in order to get it into a form that shows nonexistence but I have not been fruitful. Is there a different perspective that I should be looking at this with? This is my first time working with a problem such as this one.

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    $\begingroup$ Do you know much field-theory? Particularly the theory of algebraic field extensions. The tools from there work here, but if that will be all Greek to you ... :-) $\endgroup$
    – Jyrki Lahtonen
    Commented Aug 1, 2022 at 19:59
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    $\begingroup$ You could use the fact that $x^4 - 2$ is the minimal polynomial for $\sqrt[4]{2}$. That means, among other things, that a fourth-order polynomial with rational coefficients only has $\sqrt[4]{2}$ as a root if it is a constant multiple of $x^4-2$. $\endgroup$
    – eyeballfrog
    Commented Aug 1, 2022 at 20:08
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    $\begingroup$ $a=0$ by applying the automorphism sending $\sqrt[4]{2}$ to $-\sqrt[4]{2}$, then the result is obvious. $\endgroup$
    – reuns
    Commented Aug 1, 2022 at 20:35
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    $\begingroup$ The minimal polynomials of $\sqrt[4]2$ and $\sqrt{1+\sqrt2}$ are $x^4-2$ and $x^4-2x^2-1$ respectively. It follows $$\sqrt[4]2\notin\mathbb Q\left(\sqrt{1+\sqrt2}\right)$$ $\endgroup$
    – Piquito
    Commented Aug 1, 2022 at 23:27

1 Answer 1

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Assume there exist, raise to the square both sides and get $$\sqrt{2} = a^2 + b^2(1+ \sqrt{2}) + 2 a b \sqrt{1+\sqrt{2}}$$

Assume first that $a b \ne 0$ and get

$$\sqrt{1+\sqrt{2}} = c + d \sqrt{2}$$ for some rational $c$, $d$, and so $$1+ \sqrt{2}= (c+d\sqrt{2})^2$$

Separating the rational and irrational parts conclude that we also have $$1- \sqrt{2}= (c- d \sqrt{2})^2$$ and that is not possible.

We have the case $a b = 0$. Clearly cannot have $b=0$, so $a=0$ and $$\sqrt[4]{2} = b \sqrt{1+ \sqrt{2}}$$ or $$\frac{1+ \sqrt{2}}{\sqrt{2}} = \frac{1}{b^2}$$ again not possible.

$\bf{Added:}$ A general statement is that incommensurable (square) roots are linearly independent. That is, if $d_i \in F$ field ( of characteristic $\ne 2$), $\sqrt{d_i} \in K\supset F$, and $\frac{\sqrt{d_i}}{\sqrt{d_j}}\not \in F$ for $i\ne j$, then $(\sqrt{d_i})_i$ are linearly independent over $F$ (in our case $F = \mathbb{Q}(\sqrt{2})$).

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