if $r$ and $s$ are nonzeros, prove that $\mathbb{Q}(\sqrt{r})=\mathbb{Q}(\sqrt{s})$ if and only if $r=t^2\cdot s$ for some $t\in \mathbb{Q}$ Using the lema:
$F(u)=F(c\cdot u)=F(c+u)$ for $c\in F$
I could prove that:
If $r=t^2\cdot s$ for some $t\in \mathbb{Q}$, then $\mathbb{Q}(\sqrt{r})=\mathbb{Q}(\sqrt{t^2\cdot s})=\mathbb{Q}(t\cdot\sqrt{s})$, and since $r$ and $s$ are notzeros, then $t\neq0$, and then exists $t^{-1}$ which can conclude that $\mathbb{Q}(t\cdot\sqrt{s})=\mathbb{Q}(\sqrt{s})$. Therefore $\mathbb{Q}(\sqrt{r})=\mathbb{Q}(\sqrt{s})$.
However I could not prove the reciprocal.

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 $a=0$.
