Prove there is no $x, y \in \mathbb Z^+ \text{ satisfying } \frac{x}{y} +\frac{y+1}{x}=4$ Prove that there is no $x, y \in \mathbb Z^+$ satisfying
$$\frac{x}{y} +\frac{y+1}{x}=4$$ 
I solved it as follows but I seek better or quicker way: 
$\text{ Assume }x, y \in \mathbb Z^+\\  1+\frac{y+1}{y}+\frac{x}{y} +\frac{y+1}{x}=1+\frac{y+1}{y}+4 \\
\Rightarrow \left(1+\frac{x}{y}\right)\left(1+\frac{y+1}{x}\right)=6+\frac{1}{y}\\
\Rightarrow (x+y)(x+y+1)=x(6y+1)\\
\Rightarrow x\mid (x+y) \;\text{ or }\; x\mid (x+y+1)\\
\Rightarrow x\mid y \;\text{ or }\; x\mid (y+1)\\
\text{Put}\; y=nx ,n \in \mathbb Z^+\;\Rightarrow\; \frac{x}{nx} +\frac{nx+1}{x}=4 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x}=4-n \;\rightarrow\;(1)\\
\text{But}\; \frac{1}{n} +\frac{1}{x} \gt 0 \;\Rightarrow\; 4-n \gt 0 \;\Rightarrow\; n \lt 4\\
\text{Also}\; \frac{1}{n},\frac{1}{x}\le 1 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x} \le 2 \;\Rightarrow\; 4-n \le 2 \;\Rightarrow\; n \ge 2\\
\;\Rightarrow\; n=2 \;\text{ or }\; 3, \;\text{substituting in eq. (1), we find no integral values for } x.\\ \text{The same for the other case.}\\
$
So is there any other better or intelligent way to get this result?
 A: Hint
$$\frac{x}{y} +\frac{y+1}{x}=4\to \frac{x}{y}+\frac{y}{x}+\frac{1}{x}=4$$
Call $x/y=t\in \Bbb Q$. 
$$t+\frac{1}{t}+\frac{1}{x}=4\to xt^2+t(1-4x)+x=0$$
By Rational Roots Theorem the candidates to be a rational root, $t$, are $\{\pm1,\pm x,\pm 1/x\}$.
Now, test every root and check the value of $x$ you get.
Can you finish?
A: Rewrite the equation: $x^2 - 4yx + y^2+y = 0\implies\triangle'=(-2y)^2-1(y^2+y)= 4y^2-y^2-y = 3y^2-y=k^2 \implies y(3y-1)=k^2$ . Observe that $\text{gcd}(y,3y-1) = 1$ since if $d = \text{gcd} \implies d \mid y, d \mid 3y-1 \implies y = md, 3y-1 = nd\implies 1 = 3y -nd= 3md - nd = d(3m-n)\implies d = 1\implies y = u^2, 3y-1 = v^2, uv = k\implies 3u^2-v^2=1$ . This is a Pell equation and it either has infinitely many solutions or no solutions at all. Please check its status via google.
A: I include a proof for the original question. It is between the first two pictures.
There are infinitely many solutions with both $x,y \leq 0.$ This comes under the heading of Vieta Jumping. Given a solution to $x^2 - 4xy + y^2 + y = 0,$ we can create new solutions by alternately taking mappings on the hyperbola,
$$  (x,y) \mapsto (x, 4x-y - 1) \; \; \; , $$
$$ (x,y) \mapsto (4x-y,y) \; \; \; .  $$
The solutions on the hyperbola branch that is (mostly) in the third quadrant begin
$$ (0,0); \;  (0,-1); \;  (-4,-1); \; (-4,-16); \;  (-60,-16); \;  (-60,-225); \; \ldots  $$
There are infinitely many rational solutions with both $x,y > 0.$
$$ \left(\frac{1}{2},\frac{1}{2}\right); \; \;  \left(\frac{3}{2},\frac{1}{2}\right); \; \; \left(\frac{3}{2},\frac{9}{2}\right); \; \;  \left(\frac{33}{2},\frac{9}{2}\right); \; \; \left(\frac{33}{2},\frac{121}{2}\right); \; \;  \left(\frac{451}{2},\frac{121}{2}\right); \; \;  \left(\frac{451}{2},\frac{1681}{2}\right); \; \; \ldots  $$
The $y$ values above are of the form $\frac{b_n^2}{2},$ where $b_{n+2} = 4 b_{n+1} - b_n \; , \;$ giving $1,3,11,41,153,571,...$
If there were any integer solutions in the first quadrant, these same mappings would take us to an integer solution with both $x,y$ fairly small obeying certain explicit inequalities.

The following diagram goes with the inequality part:
If there were an integer solution with $x,y > 0$ and $x+y \geq 10,$ we are going to show that one of the mappings given reduces $x+y$ by at least two; therefore a finite number of steps would take us to an integer solution with $x,y > 0$ and $x+y < 10.$ One can quickly inspect and find that there are no integer solutions with such small numbers. So that is it.
If $x+y \geq 10$ and both positive, one case, closer to the positive $x$ axis, has $x > y$ and $y > 2.$ From the quadratic formula and the inequalities we find $x = 2y + \sqrt{3y^2 - y}.$ Note that $3y^2 - y > 1.$ At this point, we have $x+y = 3y + \sqrt{3y^2 - y}.$ Now, after we apply the mapping $ \color{magenta}{ (x,y) \mapsto (4x-y,y) \; \; \; },  $ the replacement value for $x+y$ is $3y - \sqrt{3y^2 - y},$ so it has shrunk by at least $2.$
The second case is closer to the positive $y$ axis, $y > x$ and $x > 2.$ We find $y = \frac{4x-1 + \sqrt{12x^2 - 8x+1}}{2}.$ After applying the other mapping $ \color{magenta}{ (x,y) \mapsto (x,4x-1-y) \; \; \; },  $ The new $y$ value is $y = \frac{4x-1 - \sqrt{12x^2 - 8x+1}}{2}.$ Therefore the sum we keep calling $x+y$ has decreased by $\sqrt{12x^2 - 8x+1}$ which is larger than $2$ when $x > 2.$ Again, we have shrunken $x+y$ by at least $2.$
That's it. There cannot be any integer solutions with $x,y > 0$ because there are none with $x+y \leq 10.$
$$  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc   \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  \bigcirc  $$

I found out how to draw line segments in desmos. One can put in a short table of points and have it connect them in order. Here is the hyperbola in the first quadrant, showing the first  four of the half integer points and how a stairway connects them, beginning an infinite process. The next point on this staircase, which goes up and to the right, would be $\left(\frac{33}{2},\frac{121}{2}\right);$

Alright, here is the hyperbola with the first few points in the third quadrant that have integer coordinates, joined up with green line segments, beginning with the origin and $(0,-1).$ The next point in this downward stairway would be $(-60,-16)$

A: For $x,y\in \Bbb Z^+$ we have  $$\frac {x}{y}+\frac {y+1}{x}=4\implies x^2-4xy+y^2+y=0\implies x=2y\pm \sqrt {3y^2-y}\implies$$  $$\implies \exists z\in \Bbb Z^+\;( z^2=3y^2-y=y(3y-1))\implies$$ $$ \implies\exists a,b \in \Bbb Z^+\;( y=a^2 \land 3y-1=b^2)\implies$$ $$\implies\exists a,b\in \Bbb Z^+\;(3a^2-1=b^2)\implies$$ $$ \implies \exists b\in \Bbb Z^+\;(b^2\equiv -1 \mod 3).$$ The 1st (displayed) line uses the Quadratic Formula.
In the 2nd line, $z\ne 0$ because $y\in \Bbb Z^+\implies y(3y-1)>0.$
In the 3rd line, $a$ and $b$ exist  because $y$ and $3y-1$ are co-prime members of $\Bbb Z^+$ and their product is the square of the positive integer $z.$
