<s>[INST] There are no rational $a, b, c $ with $a+b\sqrt{u}+c\sqrt{v} =\sqrt{uv} $ If $u$ and  $v$
are rationals such that
 $\sqrt{u}, \sqrt{v}, $ and $\sqrt{uv}$
are all irrational,
show that there are no
rational $a, b, $ and $c$
such that
$a+b\sqrt{u}+c\sqrt{v}
=\sqrt{uv}
$.
This is a generalization
of my answer 
to this question:
Proof by contradiction: finding integers that satisfy $a+b\sqrt{2}+c\sqrt{3}=\sqrt{6}$.
I will post a solution
in two days
if there are 
no posted solutions.
 [/INST] Lemma:  $\sqrt{u}$ and $\sqrt{v}$ must be linearly independent over the rationals.
Else $a \sqrt{u} + b \sqrt{v} \in \mathbb{Q}$ with $a,b \in \mathbb{Q} \setminus \{0\}$ $\implies$ $(a \sqrt{u} + b \sqrt{v})^2 = a^2 u + b^2 v + 2ab \sqrt{uv} \in \mathbb{Q}$ $\implies$ $\sqrt{uv} \in \mathbb{Q}$, contrary to the hypothesis.
Proof:  
$$
\sqrt{uv}=a+b\sqrt{u}+c\sqrt{v} \quad\quad \Big| \cdot \sqrt{u}
$$
$$
\begin{align}
u \sqrt{v} & =a\sqrt{u}+bu+c\sqrt{uv} \\
 & = a\sqrt{u}+bu+c(a+b\sqrt{u}+c\sqrt{v}) \\
 & = (a+bc)\sqrt{u} + c^2 \sqrt{v} + ac + bu
\end{align}
$$
By the linear independence proved in the lemma, the coefficients of $\sqrt{v}$ must match i.e. $\,u = c^2\,$ but that contradicts the irrationality of $\sqrt{u}$.
</s>