3
$\begingroup$

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?

  • 1
    Oops, I meant "6th-power primitive", or the terms do not have a 6th power $u^6$ where $u\ne1$ in common.2013-05-10

1 Answers 1

2

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. $

  • 0
    Can you answer one question of this [MO post](http://mathoverflow.net/questions/203923/)? It is similar, but focuses only on $k=2,3$.2015-04-27