Let $n$ be a positive integer. Prove that $x+y+\frac{1}{x}+\frac{1}{y}=3n$ doesn't accept solutions in the set of positive rational numbers. 
Let $n$ be a positive integer. Prove that $$x+y+\frac{1}{x}+\frac{1}{y}=3n$$
doesn't accept solutions in the set of positive rational numbers.


$\underline{\textbf{My approach:}}$ We write the expression as $$\frac{(x y+1)(x+y)}{x y}=3 n$$ We assume the contrary that $x,y$ can be rational. Let $$x=\frac{p_1}{q_1}, y=\frac{p_2}{q_2}$$ where $gcd(p_1,q_1)=gcd(p_2,q_2)=1$. Now we replace the values if $x,y$ in the expression to get $$\left(\frac{p_1 q_2+p_2 q_1}{q_1 q_2}\right)\left(1+\frac{q_1 q_2}{p_1 p_2}\right)=3 n$$ Now, I feel somehow I have to use the fact that $p_1,q_1$ and $p_2,q_2$ are coprime but I can't figure out how. Please don't give a solution, I'm only looking for some approach/hint/motivation that will help me progress. Thank you
 A: To get started, consider
$$ A  = \{ r \in \mathbb{Q} | r = q + \frac{1}{q}, q \in \mathbb{Q} \}, B = \{ n \in \mathbb{N} | n = q + \frac{1}{q} + p + \frac{1}{p}, p, q \in \mathbb{Q}\}.$$
First, list out some explicit terms of both sets.
For $A$, what terms of the form $\frac{n}{2}$ can we get? What about $ \frac{n}{3}$, $\frac{ 2n+1}{4}$, etc?
Then, try to find a description between terms of these sets other than "B is the set of integers that is the sum of 2 terms in A".

This hint will give away the entire problem, so read with caution.
It is the "description between terms of these sets".

 Show that $n \in B$ if and only if $n$ is the product of 2 terms in $A$.


 So it remains to show that there is no $\frac{p}{q} \in A$ with $ \gcd(p, q ) = 1$ and $ 3 \mid p$.


Why would we consider this / How to motivate this?

*

*Studying these sets make some sense.

*It is clear that $ 4 \in B, 5 \in B$, so $B$ is non-empty, and we would want to hunt down what these values could be.

*It is hard to prove what values are not in $B$ without some kind of classification. EG How can we show that $6, 7, 8, 9 \not \in B$?

*OP came close to showing something like $ x + y + \frac{1}{x} + \frac{1}{y} = ( a + \frac{1}{a} ) ( b + \frac{1}{b} )$, though just needed to hunt down those variables further. (Basically $ab = x, a/b = y$, but has to be justified why $xy = a^2, x/y = b^2$)

A: I tried a more number-theoretical approach.
Assumed there are solutions in the nonnegative rationals.
Let $x := \frac{x_1}{x_2}, y := \frac{y_1}{y_2} \text{ where } x_1,x_2,y_1,y_2\in\mathbb Z\setminus\{0\}$. We may also assume that
$gcd(x_1, x_2) = gcd(y_1,y_2) = 1$.
Now multiply all denominators, so we get
$$x_1^2y_1y_2 + x_1x_2y_1^2 + x_2^2y_1y_2 + x_1x_2y_2^2 = 3nx_1x_2y_1y_2 =: 3m, m\in\mathbb Z.$$
Rearranged:
$$\left(x_1^2+x_2^2\right)y_1y_2 + \left(y_1^2+y_2^2\right)x_1x_2 = 3nx_1x_2y_1y_2.$$
The most simple case is $x_1\equiv x_2\equiv 0\pmod 3$, but this contradicts since they are assumed coprime.
Furthermore, since $gcd(x_1x_2, x_1^2 + x_2^2) = gcd(y_1y_2, y_1^2 + y_2^2) = 1$ we can conclude $x_1x_2 | y_1y_2$ and vice versa. So, $x_1x_2 = y_1y_2$.
We then have $x_1^2 + x_2^2 + y_1^2 + y_2^2 = 3nx_1x_2$.
So, the sum of squares must be divisible by $3$.
This is true if and only if all four numbers are divisible (contradiction to being coprime) or exactly one of them (contradiction to $x_1x_2 = y_1y_2$). All in all, contradicton. qed
