Elementary argument for $\sqrt{2+\sqrt{2}} \not\in \mathbb{Q}[\sqrt 2]$ and possible generalization

Question

We know the minimal polynomial $\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}$ is $x^4-4x^2+2$ which by Eisenstein is irreducible, thus $\left|\mathbb{Q} \left[\sqrt{2+\sqrt{2}}\right]:\mathbb{Q}\right| = 4$, then by tower law $$\left|\mathbb{Q}\left[\sqrt{2+\sqrt{2}}\right]:\mathbb{Q}\right| =\left|\mathbb{Q}\left[\sqrt{2+\sqrt{2}}\right]:\mathbb{Q}[\sqrt{2}]\right| \Big|\mathbb{Q}[\sqrt 2] : \mathbb{Q}\Big|$$ which shows $\left|\left[\sqrt{2+\sqrt{2}}\right]:\mathbb{Q}[\sqrt{2}]\right| = 2$.

Is there an elemetary way of showing $$\sqrt{2+\sqrt{2}} \neq p + q\sqrt 2 \;\text{ for } p,q \in \mathbb{Q}?$$

And generally, is it true that $\sqrt{a+\sqrt{b}} \not\in \mathbb{Q}[\sqrt b]$ for (let us assume) $a\geq 0$ and $b$ is prime?

Answer 1

Supposing otherwise, this implies that $\sqrt{2+\sqrt{2}}=p+q\sqrt{2}$. for some $p,q\in \mathbb{Q}$. This gives:

$$2+\sqrt{2}=p^2+2pq\sqrt{2}+2q^2,$$

or,

$$\sqrt{2}=\frac{p^2+2q^2-2}{1-2pq}=\frac{a}{b},$$

where $a,b$ are integers. Note that if $1-2pq=0$, then from the above this implies that $2+\sqrt{2} = p^2+\sqrt{2}+2q^2$, or $2 = p^2+2q^2$. This implies $2|p^2$, which is ludicrous since $\sqrt{2}$ is irrational and $2q^2>0$, so $p\geq 2$.

But this is impossible, as $\sqrt{2}$ is irrational.

More generally,

$$\sqrt{b} = \frac{p^2+q^2b-a}{1-2pq}=a’/b’,$$

and again from the fact that $\sqrt{b}$ is irrational for non-square $b$, the result follows. You'll have to again double check that 1-2pq\neq 0$ by similar methods.

Answer 2

If $$\sqrt{2+\sqrt 2}=p+q\sqrt 2, $$ then by squaring, $$2+\sqrt 2=p^2+2q^2+2pq\sqrt 2 $$ so that by comparison of coefficients, $$p^2+2q^2=2,\qquad 2pq=1. $$ Writing $p=\frac ac$, $q=\frac bd$ with $a,b\in\Bbb Z$, $c,d\in\Bbb N$, $\gcd(a,c)=\gcd(b,d)=1$, $$ a^2d^2+2b^2c^2=2c^2d^2,\qquad 2ab=cd.$$ From the latter and $\gcd(a,c)=1$, we have $a\mid d$. Similarly, $b\mid c$. Then $$ 1=\gcd(a^2d^2,b)=\gcd(2c^2d^2-2b^2c^2,b)=|b|,$$ so $b=\pm1$. Similarly, $$ \{1,2\}\ni\gcd(a^2,2)=\gcd(a^2,2b^2c^2)=\gcd(a^2,2c^2d^2-a^2d^2)=a^2\ne2,$$ so that $a=\pm1$. Then $$d^2+2c^2=2c^2d^2,\qquad 2=cd, $$and we arrive at a contradiction as we need only check $c=1, d=2$ and $c=2, d=1$.

Answer 3

Here is an alternative way to finish the solution. We derive $p^2+2q^2=2,2pq=1$ as in Hagen's answer. Adding them together, we get $(p+q)^2+q^2=3$. However, $3$ is not a sum of two rational squares, which can be seen by clearing the denominators and observing that $a^2+b^2=3c^2$ can't have a nonzero integer solution, as then $a,b$ are divisible by $3$ (as we can see by considering squares modulo $3$), hence so is $c$ and we can always find a smaller solution.