Injectivity of $f:\mathbb R_{>0} \times \mathbb R \to \mathbb R^2, (x,y) \mapsto (xy, x^2 - y^2)$ I want to show that the function $f:\mathbb R_{>0} \times \mathbb R \to \mathbb R^2, (x,y) \mapsto (xy, x^2 - y^2)$ is injective. I showed this by calculating an inverse $f^{-1}:\mathrm{Im}(f) \to \mathbb R_{>0} \times \mathbb R$. 
But this involves a long calculation and nasty substitutions. So I wondered if there is a more elegant way of doing this. I would appreciate some hints :)
 A: A "geometric" answer.
Note that all points $x,y$ that get mapped into $u,t$ with the same $t$ are points from the same half circumference (centred on the origin) in the right half of the cartesian plane, with $t$ being the square of the radius $\sqrt{x^2+y^2}$. For any of these points, $u$ corresponds to twice the area $xy/2$ of the triangle inscribed in the quarter circumference (with one point in the origin and the other on the $y$ axis), with the same sign as its $y$ coordinate on the circumference. Given that there are at most two such triangles for any given area, and one gets mapped into $u>0$ and one into $u<0$ (or possibly a single degenerate one that gets mapped into $u=0$), you have your injectivity with a lot of words but few numbers!
A: Hint
Take the standard approach:
$$f(p,q)=f(m,n) → (pq,p^2-q^2)=(mn, m^2-n^2)$$
So,
$$pq=mn \quad \text{and} \quad p^2-q^2=m^2-n^2$$
$$p^2-\left(\frac{mn}{p}\right)^2=m^2-n^2 → p^4-p^2(m^2-n^2)-(mn)^2=0$$
$$\Delta=(m^2-n^2)^2+4(mn)^2=(m^2+n^2)^2$$
$$p^2=\frac{(m^2-n^2)\pm (m^2+n^2) }{2} $$
You have to prove that $p=m$ and $q=n$.
Can you finish?
A: Suppose $xy=p, x^2 -y^2=q$ Then $y=\frac p x$ and, replacind in the second equality: $x^2 - \frac {p^2}{x^2}=q$ 
Now using notation $x^2=t$ we get $t^2 -qt -p^2=0$ which, according to Vieta, has at most one positive solution. Therefore there can be at most one $x^2$ and, because $x \gt 0$, at most one $x$. From $y=\frac p x$ , we'll get at most one $y$ for that $x$.
