For what triplets (x, y, n) of positive integers is the equation
(x−y)^ n = xy ?
First n=1. Then x-y = x*y. This can be rearranged as (x+1)*(y-1) = -1. There are then two integer solutions (x,y) = (0,0) and (-2,2), but these both fail the postive requirement.
Now n=2. Then (x-y)^2 = x*y. Adding 4xy to each side and rearrange to get (x+y)^2 = 5*x*y. Then x*y and 5*x*y are both perfect squares. But that can only happen when xy=0. So the only solution in this case is (x,y)=(0,0) and that also fails the positive requirement.
Now for n>2, the 'large' values of n. Let g be the GCD of x and y. Then let a and b be the numbers so that x=a*g and y=b*g. a, b, and a-b will all be coprime to each other
Substitute the expressions in the equation to yield g^(n-2) * (a-b)^n = a*b. a-b must be a factor of the right side, but the right side is coprime to a-b. This can only happen when a-b=1.
Substitute a=b+1. Then the equation becomes g^(n-2) = b*(b+1). If n=3 then g = b*(b+1), which forms the parameterization (x,y) = (b*(b+1)^2, b^2*(b+1)). This describes the set Charlie found in his numeric search.
If g>3 then g^(n-2) is a perfect power. Since b and b+1 are coprime, they would need to both be perfect powers of n-2 for the equation to be solvable in integers. However there are no positive consecutive perfect powers so no positive solutions exist for n>3. (Aside from the trivial x=y=0 which fails the positive requirement.)
In summary, all positive integer triplets (x,y,n) which satisfy the equation are (b*(b+1)^2, b^2*(b+1), 3) for all positive integer b.