If $r$ and $s$ are nonzero, prove that $\mathbb{Q}(\sqrt{r})=\mathbb{Q}(\sqrt{s})$ if and only if $r=t^2s$ for some $t\in \mathbb{Q}$.

Question

Using the lemma $F(u)=F(cu)=F(c+u)$ for $c\in F$, I can prove:

If $r=t^2s$ for some $t\in \mathbb{Q}$, then $\mathbb{Q}(\sqrt{r})=\mathbb{Q}(\sqrt{t^2s})=\mathbb{Q}(t\sqrt{s})$, and since $r$ and $s$ are not zero, then $t\neq0$, and there exists $t^{-1}$. We can conclude that $\mathbb{Q}(t\sqrt{s})=\mathbb{Q}(\sqrt{s})$. Therefore $\mathbb{Q}(\sqrt{r})=\mathbb{Q}(\sqrt{s})$. However I can't prove the converse.

Answer 1

The statement is clear if $\sqrt{r}$ is rational: also $\sqrt{s}$ must be, so they're both squares and so is their ratio.

Assume $\sqrt{r}$ is irrational.

Since $\sqrt{s}\in\mathbb{Q}(\sqrt{r})$, there exist $a,b\in\mathbb{Q}$ such that $$ (a+b\sqrt{r})^2=s $$ Thus $a^2+rb^2+2ab\sqrt{r}=s$. Suppose $ab\ne0$; then $$ \sqrt{r}=\frac{s-a^2-rb^2}{2ab}\in\mathbb{Q} $$ a contradiction. Therefore either $a=0$ or $b=0$. The case $b=0$ is ruled out, because $\sqrt{s}$ cannot be rational. Thus $a=0$.

Answer 2

Suppose $\sqrt r\in \Bbb Q(\sqrt s)$. That means $\sqrt r=a+b\sqrt s$ for some $a, b\in \Bbb Q$.
Squaring it yields $r=a^2+2ab\sqrt s+b^2s$.

If $\sqrt s\notin \Bbb Q$, we must have $ab=0$, thus either $b=0$ when $r=a^2$ so $\Bbb Q(\sqrt r)=\Bbb Q\ne\Bbb Q(\sqrt s)$, or $a=0$ which gives the desired statement.