If $(m,n)\in\mathbb Z_+^2$ satisfies $3m^2+m = 4n^2+n$ then $(m-n)$ is a perfect square. I came across this question on another forum. The question is:

$$ \text{If $m,n\in \mathbb{Z}_+$ such that $3m^2+m=4n^2+n$, then $(m-n)$ is a perfect square.}$$

I have managed to partially prove this using this question as motivation as follows.
Let $m>n$ and $k^2 = m-n$. The problem then becomes to show $k$ is an integer. Making the substitution $m=n+k^2$ we get
$$3(n+k^2)^2+(n+k^2) = 4n^2+n$$
And solving for $n$ yields
$$n = 3k^2\pm |k|\sqrt{12k^2+1}$$
So $n$ will be an integer if and only if $12k^2+1$ is a perfect square. This is where the previous question comes in. We want all solutions $(k,N)$ to $12k^2+1=N^2$, i.e. 
$$N^2-12k^2=1$$
Using Pell's equation and Wikipedia (Pell Equation) as a guide we find the fundamental solution as $y_1=k=2, x_1=N=7$, and hence all other solutions are $x_i, y_i$ where
$$x_i+y_i\sqrt{12} = \left(7+2\sqrt{12}\right)^i.$$
It is not hard to see $y_i$ is an integer for all $i$. My conclusion is then: If $(m,n)$ is a solution then $k^2=(m-n)\in S=\{y_i^2\}_{i=1}^{\infty} = \{2^2, 28^2, 390^2,...\}$.


My questions are:
$\ \ \ \bullet$ I made the assumption that $m>n$, is this easy to show?
$\ \ \ \bullet$ If $y\in S$, is there always a solution $(m,n)$ with $(m-n)=y$ ?
$\ \ \ \bullet$ More importantly: Is there an easier way to prove this?

 A: Rewrite the original equation $3m^2+m=4n^2+n$ as 
$$12m^2+12n^2+m-n-24mn=16n^2+9m^2-24mn.$$
This factors as
$$(m-n)(12(m-n)+1)=(4n-3m)^2.$$
Since $\gcd(m-n,12(m-n)+1)=1$, it follows that $m-n$ is a perfect square, as desired.
A: Since I criticized your solution, I felt kind of obliged to provide one. Besides, it is an interesting problem. So, here it is.
What I want to show is that if $(m,n)$ is an integer solution to your equation, and $(m^*,n^*)$ is the next solution (we will order all solutions), then
$$\sqrt{m^*-n^*}=\frac{m+n}{\sqrt{m-n}} \tag{*}$$
so that, by induction, if $\sqrt{m-n}$ is an integer, $\sqrt{m^*-n^*}$ is rational and its square is integral, hence, it is an integer as well. The first non-trivial solution $m=30$, $n=26$ (see below) gives $\sqrt{30-26}=2$.
Step 0. Note that using the equation from the problem statement,
$$(m+n)^2=2(m^2+n^2)-(m-n)^2=2(m-n)(7m+7n+2)-(m-n)^2$$
so that in order to show (*) we need to show that
$$m^*-n^*=\frac{(m+n)^2}{m-n}=13m+15n+4 \tag{**}$$
Now, this is a pretty straightforward exercise.
Step 1. Pell's equation. We rewrite equation so that it looks more like a Pell's equation:
$$3(m+1/6)^2-(2n+1/4)^2=1/48$$
or, by multiplying to make all coefficients integral,
$$(12m+2)^2-3(8n+1)^2=x^2-3y^2=1$$
Step 2. Solve Pell's equation. The initial solution corresponding to $m=n=0$ is $(x,y)=(2,1)$. So, others are given by recursion:
$$x'=2x+3y,y'=x+2y$$
We need to filter out those that give non-integer values for $m$ and $n$. The chain of $(x\mod 12,y\mod 8)$ starting from the first solution: $(2,1)\rightarrow(7,4)\rightarrow(2,7)\rightarrow(1,0)\rightarrow(2,1)\rightarrow\dots$. So, the solutions $(x,y)$ giving integer $m$ and $n$ are exactly 4 steps away from each other.
$$\left(\begin{array}{} x^* \\ y^* \end{array}\right)=\left(\begin{array}{} x'''' \\ y'''' \end{array}\right)=\left(\begin{array}{} 2 & 3 \\ 1 & 2 \end{array}\right)^4\left(\begin{array}{} x \\ y \end{array}\right)=\left(\begin{array}{} 97 & 168 \\ 56 & 97 \end{array}\right)\left(\begin{array}{} x \\ y \end{array}\right)$$
And, from here, we obtain (**):
$$24(m^*-n^*)=2(x^*-2)-3(y^*-1)=2x^*-3y^*-1=26x+45y-1=$$
$$=312m+52+360y+45-1=24(13m+15n+4)$$
A: It became interesting for the General case. When the difference is a square?
Write so equation:
$$aX^2+X=bY^2+Y$$
If you use the solutions of the Pell equation.
$$p^2-abs^2=\pm1$$
Then decisions can be recorded.
$$X=\pm(p+bs)s$$
$$Y=\pm(p+as)s$$
$p,s$ - can be of any sign. So the difference will be equal.
$$X-Y=\pm(b-a)s^2$$
Mean difference solutions of the square when the difference of the coefficients of the square.
