How do you solve the Problem below?

Question

Let $u,v,w\in \mathbb{Z}>0$ denote 3 relatively prime integers(Pairwise coprime). If $(mn)$ is irrational, can we find 2 non-zero coprime (non-square) integers $u,v$ such that: $$\dfrac{\sqrt{mn}-n}{\sqrt{u}} $$ and $$ \dfrac{\sqrt{mn}-m}{\sqrt{v}} $$ are integers.

Answer 1

Hint: If $a,b$ are integers such that $$ a = \dfrac{\sqrt{2mn}-n}{\sqrt{x}} \qquad \text{and} \qquad b = \dfrac{\sqrt{2mn}-m}{\sqrt{y}} $$ it follows that $$ a \sqrt{x} + n = b \sqrt{y} + m $$ Can you proceed from here?

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Note that up to substituting $x$ and $y$ with $a^2x$ and $b^2y$, respectively, we may assume $a,b \in \{\pm 1\}$. Indeed, it isn't hard to prove that $\sqrt{2mn} \neq n$, so $a \neq 0$, and similarly for $b$. Working out my hint would then suggest $x = v$ and $y = w$, which is why I posted it.

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After this I played a bit with $(u,v,w) = (2,3,5)$ and I don't think that this problem has a solution in general, though. The issue is this: according to Wolfram Alpha we have $$ \begin{align} \sqrt{2 \, (\sqrt{2}+\sqrt{3}) \, (\sqrt{2}+\sqrt{5})} &\approx 4.79265 \\ \sqrt{2}+\sqrt{5} &\approx 3.65028 \end{align} $$ (see this and this) and $$ \alpha := \sqrt{2 \, (\sqrt{2}+\sqrt{3}) \, (\sqrt{2}+\sqrt{5})} - \sqrt{2} - \sqrt{5} \approx 1.14236 $$ The problem is that $1 = \sqrt{1} < \alpha < \sqrt{2} \approx 1.41421$, hence there is no integer $x \in \Bbb{Z}$ such that $\sqrt{x} = \alpha$.