Proof that 1/x + 1/y is distinct for distinct unordered pairs of (x,y), xy = k.

Question

Take xy = k, for nonzero k. There are many (x,y) that can satisfy this. However, how do I prove that the sums of the members of any two distinct, unordered pairs, is distinct? (This is an equivalent proof to the title question).

Answer 1

To elaborate on the discussion in the comments: suppose that $x+y=S$ and $xy=k$ are given. We claim that this data specifies the pair $(x,y)$ up to order.

Indeed, declaring $x$ to be the larger of the two we easily see that $$2x=\sqrt {S^2-4k}+S$$.

Just to emphasize, if $(x',y')$ was a another pair with $x'≥y',x'+y'=S,x'y'=k$ then the same algebra would show that $$2x'=\sqrt{S^2-4k}+S=2x\implies x'=x$$