Showing that $(x,y)=(4,5), (5,4)$ are the only positive integer solutions to $x+y=3n, x^2+y^2-xy=7n$ 
Show that $(x,y)=(4,5), (5,4)$ are the only positive integer solutions to $x+y=3n, x^2+y^2-xy=7n.$

I'm not very certain how to proceed on this problem. I know $x^2+y^2=(x+y)^2-2xy,$ so we essentially have $x+y=3n, (x+y)^2-3xy=7n$ for positive integers $x, y, n.$ However, this doesn't really help. I've also tried writing it as a fraction and doing some algebraic manipulations, but I haven't gotten anywhere either. May I have some help? Thanks in advance.

Because $x$ and $y$ are exchangeable, I represent them as
$$ x=m+d \qquad y=m-d
$$
Then
$$ x+y=3n \qquad \to \qquad 2m = 3n  \qquad \to \qquad 7n= 14/3m \tag 1$$
$$ x^2+y^2-xy = 7n \qquad \to \\
  2m^2+2d^2 - (m^2-d^2)= 7n \qquad\to \\ 
   m^2+3d^2 = 14/3m \qquad\to \tag 2 $$
$$   m^2-14/3m + (7/3)^2 = (7/3)^2-3d^2 \qquad\to $$
$$   m= 7/3 \pm \sqrt{ (7/3)^2-3d^2}  \tag 3 $$
The integer resp half-integer values of the term $7/3 \pm \sqrt{ (7/3)^2-3d^2}$ can be enumerated for $d \in \{0,1/2,1,3/2,...\}$ and all for $d \gt 1 $ become imaginary. From that only $d=0$ and $d=1/2$ are integer or half integer and lead to the solutions:
d    m                x=m+d  y=m-d
--------------------------------------- for 7/3 + sqrt(...)
0    4.66666666667    4+2/3  4+2/3
0.5  4.50000000000    5      4           <---- the single integral solution
1    3.89680525327    <fractional> 
--------------------------------------- for 7/3 - sqrt(...)
0    0                  "trivial"
0.5  0.166666666667   <fractional>
1    0.769861413392   <fractional>


Conclusion: The only integer-solutions are $(x,y)=(y,x)=(5,4)$
