# What is the sixth Martin quadruple?

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 primitive, i.e. GCD(a,b,c,d) = 1, and positive have the curious sums,

\begin{aligned} &10 + 13 + 14 + 44 = 3^4\\ &54 + 109 + 202 + 260 = 5^4\\ &102 + 130 + 234 + 318 = 2^4\cdot7^2\\ &198 + 630 + 1594 + 1674 = 2^{12}\\ &570 + 742 + 1094 + 1690 = 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?

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References to previous work? –  Gerry Myerson Sep 18 '12 at 6:52
How do the last three qualify as primitive? $1\neq \gcd (102,130,234,318)=\gcd (198,630,1594,1674)=\gcd (570,742,1094,1690)=2$ –  cyclochaotic May 9 '13 at 12:39
Oops, I meant "6th-power primitive", or the terms do not have a 6th power $u^6$ where $u\ne1$ in common. –  Tito Piezas III May 10 '13 at 19:26