In the following equation $\rho(x,y)$ returns a constant value for a given coordinate. $\mathbf n$ is the normal vector to the surface of the form $[P,Q,-1]$ and $s$ is a direction vector. $$I(x,y)=\rho(x,y)\frac{\mathbf n\cdot\mathbf s}{|\mathbf n||\mathbf s|}.$$
Using $s = [S_x,S_y,S_z]$, the equation above can be rewritten as: $$I(x,y)=\rho\frac{S_xP+S_yQ-S_z}{\sqrt{P^2+Q^2+(-1)^2}\sqrt{S_x^2+S_y^2+S_z^2}}.$$ We know that $P = -dz/ dx$ and $Q = -dz/dy$.
A single measurement of $I(x,y)$ for a given value of $s$ produces a conic section in p-q space.
Plot in the p-q plane for a constant s but different values for I(x,y) produces conic sections as shown below:
Given two measurements of I(x,y) : I1 and I2, and two corresponding values for s, S1 and S2. This might produce an intersection of conic sections in the p-q space.
I need to prove that there are at most two solutions for p and q.
This is part of a lecture for photometric stereo. The full pdf of the lecture can be found here
The full story is that two different (and nondegenerate) conics will always intersect in exactly four points, counting multiplicity and accepting points in the whole projective plane with complex coordinates. On a more mundane level, I’m sure that you can imagine two ellipses intersecting in four distinct visible points. So there must be some additional conditions forcing the existence of only two real points.