General solution of equation with coefficients the symmetric polynomials If $a,b,c$ are fixed integers, how do you find the general solution of
$$X(abc)+Y(ab+bc+ca)+Z(a+b+c)=0$$
in integers $X,Y,Z$?
 A: You may call $A=abc$, $B=ab+bc+ca$ and $C=a+b+c$, and study the equation $AX+BY+CZ=0$.
(1) $A=B=C=0$ the solutions are $(x,y,z)$ for any integer numbers $x,y,z$.
(2) $A=B=0$, and $C\neq0$, the equation is of type $CZ=0$, whose solutions are $(x,y,0)$ for any integer numbers $x,y$
If at least two of $A,B,C$ are non--zero we assume they are coprime.
(3) If $A=0$, $B,C\neq0$, the equation is of type $BY+CZ=0$, whose solutions are $(-Ck,Bk)$, for any integer numbers $x,k$.
(4) If $A,B,C\neq0$, and one solution is $(x,y,0)$, then $(x,y)$ is a solution of $AX+BY=0$, whose solutions are well known: they are $(-B'k,A'k)$, where $A'=A/\mathrm{gcd}\{A,B\}$, and $B'=B/\mathrm{gcd}\{A,B\}$.  The same analysis can be realized if either $x=0$ or $y=0$.
If $(x,y,z)$ is a solution and $x,y,z\neq0$ then $Ax+By+C=0$, where we assume $z=1$, hence $(x,y)$ is a solution of $AX+BY=-C$, with well known solutions, if they exist, i.e., if $\mathrm{gcd}\{A,B\}\mid{C}$, and since $\mathrm{gcd}\{A,B,C\}=1$ if $\mathrm{gcd}\{A,B\}=1$. General solution is $(xk,yk,k)$, for any integer number $k$. The same analysis can be realized for $x$ and $y$. 
