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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?

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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
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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

1 Answer 1

up vote 2 down vote accepted

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 $$ \{a,b,c,d\}=\{259, 1307, 3485, 9349 \}. $$


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

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Oleg567: Thanks for answering this old question. Now why I am not surprised that its linear sum $2^4 3^6$ are powers of small primes? What is it with these quadruples? –  Tito Piezas III Apr 24 at 14:30
    
Tito: hmm, I have no intuition why linear sums are powers of small primes. 7th quadruple has product of 3 smallest primes. –  Oleg567 Apr 25 at 3:38
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The phenomenon does not extend to,$$a^2+b^2+c^2+d^2 = y^2$$ $$a^4+b^4+c^4+d^4 = z^4$$ the smallest solution of which is $a,b,c,d = 15935,\,27022,\, 57910,\, 59260$ (See this post.) The linear sum is not a smooth number. –  Tito Piezas III Apr 25 at 16:04
    
Can you answer one question of this MO post? It is similar, but focuses only on $k=2,3$. –  Tito Piezas III Apr 27 at 7:58

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