Diophantine equation, 3 variables How do I solve the following equation, where $x,y,z$ have to be positive integers?
$$ \frac{x^2}{y} + \frac{y^2}{z}+ \frac{z^2}{x}= \frac{y^2}{x} + \frac{z^2}{y} + \frac{x^2}{z} $$
Given that $$xyz=2205$$
 A: Multiply both sides of the first equation by $xyz$ and try to factor it to linear expressions, you will get$$(x-y)(x-z)(y-z)(x+y+z)=0$$And notice that $2205=3^2 \times 5 \times 7^2$, so exactly two of the three variables are equal.
A: Note that if $x=y$, then the equation holds for any $x,z$ satisfying $x^2z=2205=3^27^25$. That gives us the solutions $(1,1,2205),(3,3,245),(7,7,45),(21,21,5)$. Symmetry then gives us $(1,2205,1),(2205,1,1),(3,245,3),(245,3,3),(7,45,7),(45,7,7),(21,5,21),(5,21,21)$.
If we multiply the original equation by $xyz$ we get $$x^3(y-z)+y^3(z-x)+z^3(x-y)=0\ (*)$$ We suspect that $x-y$ must be a factor given the solutions above, so we rearrange $x^3(y-z)+y^3(z-x)$ as $z(y^3-x^3)+xy(x^2-y^2)$ $=(y-x)(z(y^2+xy+x^2)-xy(x+y))$. 
Hence $(*)$ becomes $(y-x)(z(x^2+xy+y^2)-xy(x+y)-z^3)$. Since it has a factor $(y-x)$ by symmetry it must also have $(z-y)$ and $(x-z)$ and we find that $z(x^2+xy+y^2)-xy(x+y)-z^3=(z-y)(x-z)(x+y+z)$. Since $x,y,z$ are positive $x+y+z\ne0$ and so two of $x,y,z$ must be equal.
So the only solutions are the 12 in the first paragraph above.
