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Given an integer solution $s_m$ to the system,

$$x_1^2+x_2^2+\dots+x_n^2 = y^2$$

$$x_1^3+x_2^3+\dots+x_n^3 = z^3$$

and define the function,

$$F(s_m) = x_1+x_2+\dots+x_n$$

For $n\geq3$, using an elliptic curve, it can be shown there are an infinite number of primitive solutions. For $n=4$, we have positive and primitive,

$$\begin{aligned} s_1 = 10,\, 13,\, 14,\, 44;\quad &F(s_1) = 3^4\\ s_2 = 54,\, 109,\, 202,\, 260; \quad &F(s_2) = 5^4\\ s_3 = 51,\, 65,\, 117,\, 159; \quad &F(s_3) = 2^3 7^2\\ s_4 = 99,\, 315,\, 797,\, 837; \quad &F(s_4) = 2^{11}\\ s_5 = 285,\, 371,\, 547,\, 845; \quad &F(s_5) = 2^{11}\\ s_6 = 815,\, 1297,\, 1781,\, 1939; \quad &F(s_6) = 2^3 3^6\\ s_7 = 259,\, 1307,\, 3485,\, 9349;\quad &F(s_7) = 2^6 3^2 5^2\\ \end{aligned}$$

with $s_6$ and $s_7$ found by MSE user oleg567.

Questions (for $n=4$):

  1. Why do these (apparently smallest) solutions have $F(s_m)$ as a smooth number?
  2. Is this really the complete list with all $x_i<10^4$? (These are the smallest if the list in this post is appropriately complete.)

Comment: For $n=3,5$, small solutions do not necessarily have $F(s_m)$ that are smooth numbers.

Edit: Using $s_1$, we have,

$$P_k(x)=(\color{blue}{10}x^2 - x + 88)^k + (\color{blue}{13}x^2 + 68x + 28)^k + (\color{blue}{14}x^2 - 68x + 26)^k + (\color{blue}{44}x^2 + x + 20)^k$$

For $k=3$, it is already a cube,

$$P_3(x) = 45^3(x^2+2)^3$$

For $k=2$, it a quartic polynomial to be made a square with a rational point, hence an elliptic curve,

$$P_2(x) = y^2$$

with initial points $x = 68,\,165/94$, etc. All of the $s_m$ can be used in a similar quadratic form identity. Thus, after removing denominators, the system has an infinite number of integer solutions.

However, we are interested in the behavior of the small positive solutions and which seem to be knowable only by semi-brute force searching.

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  • $\begingroup$ [there are an infinite number of primitive solutions]-->what do you mean by "primitive" ? $\endgroup$
    – Duchamp Gérard H. E.
    Apr 26, 2015 at 4:15
  • $\begingroup$ If I didn't make any mistakes, this system of equations defines a rationally connected 3-dimensional projective variety. There's a decent chance that we can parametrize the rational solutions. $\endgroup$
    – zeb
    Apr 26, 2015 at 4:20
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    $\begingroup$ If remove restriction that $F(s_m)$ must be a perfect square (as in MSE), then list of $s_m$ for $k=2,3$ is much wider (sorted by $F(s_m)$): $$ \begin{array}{rl} s_1=\{10,13,14,44\};&F(s_1)=81=3^4\\ s_2=\{51,65,117,159\};&F(s_2)=392=2^3 7^2\\ s_3=\{54,109,202,260\};&F(s_3)=625=5^4\\ s_4=\{62,399,402,624\};&F(s_4)=1487\\ s_5=\{140,346,417,754\};&F(s_5)=1657\\ s_6=\{217,330,496,626\};&F(s_6)=1669\\ s_7=\{39,341,447,933\};&F(s_7)=1760=2^5\cdot 5\cdot11\\ s_8=\{104,399,632,760\};&F(s_8)=1895=5\cdot 379\\ s_9=\{19,117,711,1065\};&F(s_9)=1912=2^3 239 \end{array} $$ $\endgroup$
    – Oleg567
    Apr 30, 2015 at 13:36
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    $\begingroup$ $$ \begin{array}{rl} s_{10}=\{78,521,658,680\};&F(s_{10})=1937=13\cdot 149\\ s_{11}=\{129,337,551,967\};&F(s_{11})=1984=2^6 31\\ s_{12}=\{285,371,547,845\};&F(s_{12})=2048=2^{11}\\ s_{13}=\{99,315,797,837\};&F(s_{13})=2048=2^{11}\\ s_{14}=\{273,353,579,939\};&F(s_{14})=2144=2^5 67\\ s_{15}=\{175,584,708,1052\};&F(s_{15})=2519=11\cdot 229\\ s_{16}=\{284,699,838,910\};&F(s_{16})=2731\\ s_{17}=\{170,659,712,1330\};&F(s_{17})=2871=3^2\cdot 11\cdot 29\\ s_{18}=\{137,410,586,1756\};&F(s_{18})=2889=3^3 107\\ s_{19}=\{220,444,903,1416\};&F(s_{19})=2983=19\cdot 157\\ \ldots \end{array} $$ $\endgroup$
    – Oleg567
    Apr 30, 2015 at 13:42
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    $\begingroup$ @Oleg567 I have the impression that using the following restriction "either $F(s_m)=q^2$ or $F(s_m)=2q^2$", there are only solutions with smooth $F(s_m)$ so far. So it might be worth looking at that modified question. $\endgroup$
    – Wolfgang
    Jun 10, 2015 at 8:14

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