Diophantine system of equations I'm trying to know if there is an efficient way to find the smallest (i.e lexicographically) triplet $(a,b,c)$ of integers verifying $$a^2+b^2+c^2 = x$$ $$a^3+b^3+c^3=y$$ $$a^4+b^4 + c^4 = z$$ if $(x,y,z)$ is known. 
We assume that a solution exists for that triplet $(x,y,z)$.
Originally, this question is part of an algorithmic problem. So what I want is a fast way to find the triplet without having to brute-force.
What I've tried (the brute-force approach): going through all the possible values of $a$ since we can deduce an upper bound for its value, we remain with three equations where it's easy to find $b$ and $c$ and see if they are integers.
EDIT: We may assume as well that $a<b<c$.
 A: If $a,b,c$ are integers such that
$$a^2 + b^2 + c^2 = x$$
$$a^3 + b^3 + c^3 = y$$
$$a^4 + b^4 + c^4 = z$$
then the following divisibility conditions must hold:
$$
\left(
a + b + c
\right)
\mid 
\left(
x^2 - 2z
\right)
$$
$$
\left(
4\left(ab + bc + ca\right)
\right) 
\mid 
\left(
x^4 + 6x^2z - 16xy^2 + 9z^2
\right)
$$
$$
\left(
2abc
\right)^2 
\mid 
\left(
x^6 - 4x^3y^2 - 3x^2z^2 + 12xy^2z - 4y^4 - 2z^3
\right)
$$
Of these, the last will cut down your search considerably.
A: If you have software to find real roots, the following result provides another way to find qualifying triples $a,b,c$ ...

Proposition: If $a,b,c$ satisfy $a^2 + b^2 + c^2 = x$, $a^3 + b^3 + c^3 = y$, $a^4 + b^4 + c^4 = z$, then $a,b,c$ are roots of the following 12'th degree polynomial:
$$
p(t) = \left(12\right)\left(t^{12}\right)
-\left(24x\right)\left(t^{10}\right)
-\left(16y\right)\left(t^9\right)
+\left(24x^2-12z\right)\left(t^8\right)
+\left(48xy\right)\left(t^7\right)
$$
$$
-\left(24x^3+8y^2\right)\left(t^6\right)
-\left(24x^2y-24yz\right)\left(t^5\right)
$$
$$
+\left(15x^4+6x^2z-24xy^2+3z^2\right)\left(t^4\right)
+\left(8x^3y-24xyz+16y^3\right)\left(t^3\right)
$$
$$
-\left(6x^5-12x^2y^2-6xz^2+12y^2z\right)\left(t^2\right)
+\left(x^6-4x^3y^2-3x^2z^2+12xy^2z-4y^4-2z^3\right)
$$
