Markov-Hurwitz equation 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.
 A: See Theorem 6.1 of http://www.math.uconn.edu/~kconrad/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.
A: 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.
A: The equation can be re-written as 
$(dx)^2+(dy)^2+(dz)^2=(dx)(dy)dz)$
i.e $(dx,dy,dz)$ are solutions to 
$a^2+b^2+c^2=abc $
In the above equation if any one of $a,b,c \equiv 0 \ $mod $3$ then the RHS is $\equiv 0 $ mod $3$. This means that all of $a,b,c$ have to be divisible by 3. If $a^2,b^2,c^2$ are $\equiv 1 $ mod $3$ then LHS will be $\equiv 0 $ mod $3$ but not RHS, a contradiction. Hence $a,b,c$ are divisible by 3. Set $a=3m, b=3n, c=3p$ we get
$m^2+n^2+p^2=3mnp$
We note that if $(m,n,p)$ is a solution then so if $(m,n,3mn-p)$. The only solutions for which either 2 or more variables are equal are $(1,1,1)$ and $(1,1,2)$(note that 2=3*1*1-1, so this solution too can be derived from $(1,1,1)$).  So if we start from $(1,1,1)$ we can show that we get a sequence of solutions for which the tuple $(m,n,p)$ are co-prime. Hence the only solutions for original equation are for d=1 and d=3
