There does not exist natural numbers such that $\frac{m}{n}+\frac{n+1}{m}=4$. 
Show that there does not exist positive integers $m$ and $n$ such that $$\frac{m}{n}+\frac{n+1}{m}=4.$$

I ended up getting that $4\mid n$ and $2\mid m$, and now I am stuck. I have no idea how to proceed. 
Any help will be appreciated.
 A: Multiply the equation $$\frac{m}{n}+\frac{n+1}{m}=4$$ 
by $mn$ to get $$m^2+n^2+n=4mn$$
This can be written as $$(m-2n)^2-3n^2+n=0$$
and this is equivalent to $$12(m-2n)^2-(6n-1)^2+1=0$$
So with $u=6n-1$ , $v=m-2n$ , we get $$u^2-12v^2=1$$
This is a Pell-equation with fundamental solution $(7/2)$ and the general solution of the Pell-equation in positive integers is $$\pmatrix{7&24\\2&7}^k\cdot \pmatrix{7\\2}$$
Modulo $6$ , the general solution is $$\pmatrix{u\\v}=\pmatrix{1&0\\2&1}^k\cdot \pmatrix{1\\2}$$
and because of $$\pmatrix{1&0\\x&1}\cdot \pmatrix{1&0\\2&1}=\pmatrix{1&0\\x+2&1}$$ we can conclude that $$u\equiv 1\mod 6$$
But this is a contradiction to $u=6n-1$. Therefore the original equation has no solution in positive integers.
A: Solving for $m$ you find $$m=2n\pm\sqrt{3n^2-n},$$ which implies that $3n^2-n=n(3n-1)$ is a square. If the product of any two coprime integers is a square then they are both squares. Let $n=a^2$, $3n-1=b^2$. Then $b^2-3a^2=-1$, which is impossible modulo $4$.
