How to solve for solutions to this diophantine? I have the diophantine equation $y(x+y+z) =  xz$ where all variables are positive integers. Given some bound $y \leq B$, how can I count the number of solutions?
 A: Notice 
$$ y(x+y+z) = xz \iff 2y^2 = (y-x)(y-z)$$
For any solution $(x,y,z) \in \mathbb{Z}_{+}^3$ of above equation, $2y^2 > 0$ implies $(y-x), (y-z)$ are either both positive or both negative. The $1^{st}$ case has been ruled out because that will imply $0 < (y-z), (y-x) < y$ and make $(y-x)(y-z) < 2y^2$.
As a result, any solution of above equation must have the form:
$$(x,y,z) = (y+d_1, y, y+d_2)\tag{*}$$
where $d_1, d_2 \in \mathbb{Z}_{+}$ are divisors of $n$ and $d_1 d_2 = 2y^2$. If we consider solutions differ in ordering of $x$ and $z$ as distinct, the number of solutions of $(*)$ for $y$ equals to a fixed $n$ is just $d(2n^2)$, the number of divisors of $2n^2$. The total number of solutions for  $0 < y \le B$ becomes:
$$\mathscr{N}_B = \sum_{k=1}^{B} d(2k^2)$$
Dirichlet has showed the average order of the divisor function satisfies an inequality:
$$\sum_{k=1}^{x} d(k) \simeq x\log x + (2\gamma - 1)x + O(\sqrt{x})$$
Since $d(2k^2) > d(k)$, this immediately give us a lower bound of number of solutions:
$$\mathscr{N}_B = \sum_{k=1}^{B} d(2k^2) \ge O(B \log B)$$
A: If
$$
y=(x-z)
$$
then
$$
2x^2 - 2xz = xz \Longrightarrow 2x^2 - 3xz = 0 
$$
Which has integer solutions:
$$
x=3n, n\in \mathbb{Z}, z= 2n, n\in \mathbb{Z} \Longrightarrow y= n\in \mathbb{Z}
$$
For example, pick any positive integer $m$. Then:
$$
(x,y,z) = (3m,m,2m) 
$$
is a solution to your equation and the number of these types of solutions under a certain bound $y\le B$ where $B$ is a positive integer is $B$. 
Of course these do not encapsulate all solutions. 
$$
(x,y,z) = (7,5,30)
$$
Is a solution that doesn't have this construction. 
We can say however that if $y \le B$ and $B  >> 1$ then $B$ is a lower bound for the number of solutions. 
A: You can rearrange your equation to $y^2+(x+z)y-xz=0$ and use the quadratic equation to get $y=\frac 12\left(-(x+z)+\sqrt{x^2+6xz+z^2}\right)$ where we took the plus sign to make $y$ positive.  Now the square root needs to be integral.  Because of symmetry, we can require that $x \lt z$ and any multiple of a solution will again be a solution.  I find a class of solutions $(n, n-1,2n^2-3n+1)$ that gives $(2,1,3), (3,2,10), (4,3,21), (5,4,36), \ldots$ and another $(2n+1,2n-1,4n^2-2n)$ giving solutions $(5,3,12),(7,5,30),(9,7,56),(11,9,90)\ldots$  It seems likely there are more.  It looks like grouping them by $x-y$ could be useful.
A: the equation:
$y(y+x+z)=xz$
Has the solutions:
$y=ps$
$x=2s^2+ps$
$z=p^2+ps$
$p,s$ - integers and sets us.
