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.}\\
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So is there any other better or intelligent way to get this result?

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.
