By symmetry if (x,y) works then (y,x) works.
If y=x+1 then
x+(x+1)+1 = 2(x+1) divides 2x(x+1) and
x+(x+1)1 = 2x divides x^{2} + (x+1)^{2} − 1 = 2x(x+1)
so all of (x,x+1) and (x,x1) works.
There appear to be other sporadic cases to each of the two divisions, but I haven't found any in common yet.

Posted by Jer
on 20150616 09:30:04 