What is the sixth Martin quadruple $\sqrt[n]{x_1^k+x_2^k+x_3^k+x_4^k} =\text{Integer}$ for $k=1,2,3$? Define a Martin quadruple {a,b,c,d} as a solution in non-zero integers to the system,
$a+b+c+d = x^2$
$a^2+b^2+c^2+d^2 = y^2$
$a^3+b^3+c^3+d^3 = z^3$
It can be shown that there are an infinite number of solutions. However, the smallest five that are positive and 6th-power primitive (no common factor that is a 6th power, re cyclochaotic's comment below) have the curious linear sums as smooth numbers,
$\begin{aligned}
&10 + 13 + 14 + 44 = 9^2 = 3^4\\
&54 + 109 + 202 + 260 = 25^2 = 5^4\\
&102 + 130 + 234 + 318 = 28^2 = 2^4\cdot7^2\\
&198 + 630 + 1594 + 1674 = 64^2 = 2^{12}\\
&570 + 742 + 1094 + 1690 = 64^2 = 2^{12}\end{aligned}$
found by James Allen and Seiji Tomita.  Of course, the squares and the cubes of the addends also add up to a square and cube, respectively.
What is the sixth such quadruple?

The sixth primitive positive one is
$$
\{a,b,c,d\}=\{1630,2594, 3562, 3878 \}.
$$

$$
1630+2594+3562+3878=108^2=2^4 3^6;
$$
$$
1630^2+2594^2+3562^2+3878^2=6092^2;
$$
$$
1630^3+2594^3+3562^3+3878^3=5004^3.
$$

(update) 
The $7$th such quadruple is 
$$
\{a,b,c,d\}=\{259, 1307, 3485, 9349 \}.
$$

$$
259+1307+3485+9349=120^2=2^6 3^2 5^2;
$$
$$
259^2+1307^2+3485^2+9349^2=10066^2;
$$
$$
259^3+1307^3+3485^3+9349^3=9516^3.
$$
