Markov-Hurwitz equation

Question

Prove that the Markov-Hurwitz equation $x^2+y^2+z^2=dxyz$ is solvable in positive integers iff d= 1 or 3. Of course the reverse direction is easy, just set x=y=z=1, d=3. But I really have no idea how to attack the forward direction.

Answer 1

See Theorem 6.1 of https://kconrad.math.uconn.edu/blurbs/ugradnumthy/descent.pdf. By the way, the point of the case d = 1 and d = 3 is not that there are integral solutions (that is trivial) but that they can be systematically derived from small obvious solutions by a recursive method. See the start of Section 6 at that link for a discussion of that issue and a reference for more information.

Answer 2

For starters, d must be odd. For if d is even, either all of x,y, and z are even or two are odd and one is even. If they are all even, there is also a solution $(x/2)^2 + (y/2)^2 + (z/2)^2 = (2d)(x/2)(y/2)(z/2)$ with smaller x,y,z. If only one of x,y,z is even and d is even, the LHS is 2 mod 4 and the RHS is 0 mod 4.