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I. Fifth Powers

The Diophantine equation,

$$x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k\tag1$$

for $k=5$ is quite well-explored. It has an infinite number of primitive solutions (as points on an elliptic curve, or an infinite family of polynomials). They can be simultaneously true for $k=1,5$ and in a 2013 paper, Choudhry and Wroblewki found an infinite subset that satisfy the side condition,

$$\sum\limits^3 x_i = \sum\limits^3 y_i = 0$$

II. Seventh Powers

Since the first solution to,

$$x_1^k+x_2^k+x_3^k+x_4^k = y_1^k+y_2^k+y_3^k+y_4^k\tag2$$

for $k=7$ was found almost 30 years ago by Randy Ekl in 1996, only $113$ primitive solutions (in positive and negative integers) have been found so far. In a 2000 paper, Choudhry found solutions simultaneously true for $k = 1,3,7$ and satisfy the analogous side condition,

$$\sum\limits^4 x_i = \sum\limits^4 y_i = 0$$

Wrobrewski would later find more solutions, for a total of 24. They used the form,

$$(X_1-X_2-X_3)^k+(-X_1+X_2-X_3)^k+(-X_1-X_2+X_3)^k+(X_1+X_2+X_3)^k = (Y_1-Y_2-Y_3)^k+(-Y_1+Y_2-Y_3)^k+(-Y_1-Y_2+Y_3)^k+(Y_1+Y_2+Y_3)^k$$

which is identically true for $k=1$ and is true for $k=3,7$ if the two conditions,

$$X_1X_2X_3=Y_1Y_2Y_3$$ $$2(X_1^4 + X_2^4 + X_3^4) - 5(X_1^2 + X_2^2 + X_3^2)^2= 2(Y_1^4 + Y_2^4 + Y_3^4) - 5(Y_1^2 + Y_2^2 + Y_3^2)^2$$

are met. Using the $x_i,y_i$ (see euler.free.fr), I derived their $X_i, Y_i$. Surprisingly, seven had a simple linear constraint, namely $X_3=2Y_3$. Let,

$$X_1,\,X_2,\,X_3 =u_1 u_2,\;u_3 u_4,\;2u_5$$ $$Y_1,\,Y_2,\,Y_3 =2u_1 u_4,\;u_2 u_3,\;u_5$$

then solutions are,

$$\begin{array}{|c|c|c|c|c|c|} \text{#}&u_1&u_2&u_3&u_4&u_5\\ 1&127 &139 &313 &1 &23647\\ 2&227 &27 &113 &13 &1305\\ 3&14737 &139 &15899 &25 &544069 \\ 4&303 &304 &338 &37 &24616\\ 5&431 &187 &365 &49 &6929\\ 6&871 &163 &4364 &127 &254405 \\ 7&439 &2459 &247 &1175 &261851 \\ \end{array}$$

Looking at $u_4$, it is tempting to speculate there is a pattern in the $u_i$ and that there are infinitely many $u_i$.

Question:

  1. The first condition $X_1X_2X_3=Y_1Y_2Y_3$ is easily met via the $u_i$. After doing so, can the quartic second condition be split as an intersection of two quadric surfaces as Elkies did for $x_1^4+x_2^4+x_3^4= 1,$ or as Bremner and Ulas did for $x_1^6+x_2^6+x_3^6= y^2$?

P.S. Eq.1 is briefly discussed in this MO post.

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  • $\begingroup$ This does not answer your questions, but upon reading the paper, I came across one minute detail. In the first solution mentioned, where $X_1 = 324, X_2 = 5439, X_3 = 893$, what is the "suitable transposition" involved? The $X_i$'s mentioned does not imply $x_1 = 1741$ without that transposition. $\endgroup$
    – Jesper Petersen
    Commented Jan 2, 2015 at 20:04
  • $\begingroup$ @JesperPetersen: These solutions have the special property that $\sum_i^4 x_i = \sum_j^4 y_j = 0$. Thus, some of terms are necessarily negative. For aesthetics, Choudhry transposed negative terms to the other side of the equation, so labelling is not rigid. For example, $X_1+X_2+X_3=324+5439+893=2\cdot3328$ where one assumes $y_4 = 3328$. But $-3328$ is really one of the $x_i$ transposed to the other side. $\endgroup$
    – Tito Piezas III
    Commented Jan 2, 2015 at 21:24
  • $\begingroup$ So if I get it right, does that mean that all solutions of Choudhry always have certain terms with equal sums, as e.g. $ 1741+ 3476= 1937+ 3280$ and $2435 + 3004 = 2111 + 3328$ in the first example? Which means he has added a heavy constraint. So he has not exactly "reduced" the original equation (2) [if true for k=1, 3, 7] to (3), but could we rather say: he was lucky that his additional constraint made a computer search more efficient, which allowed him to find a bunch of solutions? $\endgroup$
    – Wolfgang
    Commented Jan 4, 2015 at 18:28
  • $\begingroup$ @Wolfgang: Yes, you are correct. I did not phrase it correctly. There are four constraints: that $k=1,3,7$ and 4th, $\sum\limits^4 x_i = \sum\limits^4 y_i = 0$. There in fact is a solution to $k=1,3,7$, namely $$u_i = 184, 443, 556, 698, -230, -353, -625, -673$$ found by Nuutti Kuosa in 1999 (before Choudhry's paper) that does not have the 4th constraint, namely there is no partition such that $\sum\limits^4 u_i = 0$. (I had checked it before.) $\endgroup$
    – Tito Piezas III
    Commented Jan 4, 2015 at 19:34
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    $\begingroup$ @Wolfgang: The thing with odd powers is it is better to move them all to one side. Thus I prefer $\sum\limits^8 u_i^k = 0$ versus $\sum\limits^4 x_i^k = \sum\limits^4 y_i^k$. The latter can be mis-leading. For example, one of the seven in the table actually is a $\sum\limits^3 x_i^k = \sum\limits^5 y_i^k$ in positive terms. $\endgroup$
    – Tito Piezas III
    Commented Jan 4, 2015 at 19:58

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