Positive Integer solutions for $(x-y)^2 = \frac{4xy}{x+y-1}$

Question

I was solving a question which asked for the number of positive integer solutions to the problem, $(x-y)^2 = \frac{4xy}{x+y-1}$. I started by finding solutions using hit and trial. First I found $(1,3)$ (and hence $(3,1)$ as the equation is symmetric in $x$ and $y$). Then I found $(3,6)$ and then $(6,10)$. Soon I observed a pattern $1,3,6,10,15,21,...$ and proved that any two consecutive terms of this series that is $(\frac{n(n+1)}{2},\frac{(n+1)(n+2)}{2})$ satisfies the equation and hence the number of solutions become infinite. My question is, is there another, more natural way to reach this solution set? In my case it seems to me I accidentally found it by making a big guess from the solutions I got from hit and trial. Also are there any other solution sets for the above equation?

Answer 1

Just try and expand ?

$$(x-y)^2(x+y-1)-4xy = (x-y)^2(x+y) - ((x-y)^2+4xy) = (x+y)(x^2-2xy+y^2-x-y) = 0.$$ Since you only want positive pair of solutions, we disregard $x+y = 0$ and then obtain: $$x_{1,2} = \dfrac{2y+1 \pm\sqrt{8y+1}}{2}$$ and then you can finish it from here.

Answer 2

Wlog, $k:=y-x>0.$ The equation is then equivalent to $$k^2(2x+k-1)=4x(x+k)$$ i.e. $$x^2-\frac{k(k-2)}2x-\frac{k^2(k-1)}4=0.$$ The roots are $x=\frac{k(k-1)}2\ge0$ and $-\frac k2<0.$ The only positive integer solutions are indeed those you found: $$x=\frac{k(k-1)}2,\quad y=\frac{k(k+1)}2,\quad k\text{ integer }\ge2.$$