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 is infinitely many $u_i$.

Question:

1. The first condition $X_1X_2X_3=Y_1Y_2Y_3$ is easily met. 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.

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.

The Diophantine equation,

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

for either $k=5$ or $6$ is quite well explored, and it has long been known that it has an infinite number of primitive solutions (as points on an elliptic curve, or an infinite family of polynomials,). However,

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

for either $k=7$ or $8$, results seem to be harder to find. Only one solution found back in 2006 is known for $k=8$. Many solutions exist when $k=7$, but it is not known if there are infinite many. Choudhry reduced $(2)$ when it is true for $k=1,3,7$ plus a fourth constraint (re Wolfgang's comment below), namely $\sum\limits^4 x_i = 0$ (and since $k=1$, then also $\sum\limits^4 y_i=0$ ), to a multi-variable cubic equation by passing through the simultaneous equations, in Choudhry's notation,

$$\begin{aligned} X_1X_2X_3\,&=Y_1Y_2Y_3\\ 2(X_1^4 + X_2^4 + X_3^4) - 5(X_1^2 + X_2^2 + X_3^2)^2\, &= 2(Y_1^4 + Y_2^4 + Y_3^4) - 5(Y_1^2 + Y_2^2 + Y_3^2)^2 \end{aligned}\tag3$$

Using that cubic, Choudhry (3) and Wroblewski (21) found a total of $3+21=24$ solutions to $(2)$ valid for $k=1,3,7$. Results can be found in euler.free.fr. Using the $x_i,y_i$, I derived their $X_i, Y_i$. Surprisingly, seven had a simple linear constraint, namely $X_3=2Y_3$. Let,

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

then solutions are,

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

It was hard to find commonalities for the other seventeen solutions.

Questions:

1. Can the simultaneous eqns $(3)$, or the cubic described by Choudhry in the paper, be reduced to an elliptic curve?
2. Why does almost $1/3$ of the $24$ known solutions have $X_3=2Y_3$? Translated into the addends $x_i,y_i$ of $(2)$, this is equivalent to the 5th constraint that $x_1+x_2 = 2y_1+2y_2$. Is there some identity behind it? For example, I found this 7th deg multi-grade,$$1 + 5^k + (3+2y)^k + (3-2y)^k + (-3+3y)^k + (-3-3y)^k = \\(-2+x)^k + (-2-x)^k + (5-y)^k + (5+y)^k$$ for $k = 1,3,5,7,\;$ if $x^2-10y^2 = 9$, guaranteeing an infinite supply with the constraint $x_2 = 5x_1$, though I doubt anything for $(2)$ would be as simple as this.

Note:

However, the fifth degree version,

$$x_1^5+x_2^5+x_3^5 = y_1^5+y_2^5+y_3^5$$

$$x_1+x_2+x_3 = y_1+y_2+y_3=0$$

does have a polynomial identity behind it, found by Choudhry and Wrobleski. (See "A quintic Diophantine equation with applications to two Diophantine systems concerning fifth powers")

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

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 is infinitely many $u_i$.

Question:

1. The first condition $X_1X_2X_3=Y_1Y_2Y_3$ is easily met. 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.

The Diophantine equation,

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

for either $k=5$ or $6$ is quite well explored, and it has long been known that it has an infinite number of primitive solutions (as points on an elliptic curve, or an infinite family of polynomials,). However,

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

for either $k=7$ or $8$, results seem to be harder to find. Only one solution found back in 2006 is known for $k=8$. Many solutions exist when $k=7$, but it is not known if there are infinite many. Choudhry reduced $(2)$ when it is true for $k=1,3,7$ plus a fourth constraint (re Wolfgang's comment below), namely $\sum\limits^4 x_i = 0$ (and since $k=1$, then also $\sum\limits^4 y_i=0$ ), to a multi-variable cubic equation by passing through the simultaneous equations, in Choudhry's notation,

$$\begin{aligned} X_1X_2X_3\,&=Y_1Y_2Y_3\\ 2(X_1^4 + X_2^4 + X_3^4) - 5(X_1^2 + X_2^2 + X_3^2)^2\, &= 2(Y_1^4 + Y_2^4 + Y_3^4) - 5(Y_1^2 + Y_2^2 + Y_3^2)^2 \end{aligned}\tag3$$

Using that cubic, Choudhry (3) and Wroblewski (21) found a total of $3+21=24$ solutions to $(2)$ valid for $k=1,3,7$. Results can be found in euler.free.fr. Using the $x_i,y_i$, I derived their $X_i, Y_i$. Surprisingly, seven had a simple linear constraint, namely $X_3=2Y_3$. Let,

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

then solutions are,

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

It was hard to find commonalities for the other seventeen solutions.

Questions:

1. Can the simultaneous eqns $(3)$, or the cubic described by Choudhry in the paper, be reduced to an elliptic curve?
2. Why does almost $1/3$ of the $24$ known solutions have $X_3=2Y_3$? Translated into the addends $x_i,y_i$ of $(2)$, this is equivalent to the 5th constraint that $x_1+x_2 = 2y_1+2y_2$. Is there some identity behind it? For example, I found this 7th deg multi-grade,$$1 + 5^k + (3+2y)^k + (3-2y)^k + (-3+3y)^k + (-3-3y)^k = \\(-2+x)^k + (-2-x)^k + (5-y)^k + (5+y)^k$$ for $k = 1,3,5,7,\;$ if $x^2-10y^2 = 9$, guaranteeing an infinite supply with the constraint $x_2 = 5x_1$, though I doubt anything for $(2)$ would be as simple as this.

Note:

However, the fifth degree version,

$$x_1^5+x_2^5+x_3^5 = y_1^5+y_2^5+y_3^5$$

$$x_1+x_2+x_3 = y_1+y_2+y_3=0$$

does have a polynomial identity behind it, found by Choudhry and Wrobleski. (See "A quintic Diophantine equation with applications to two Diophantine systems concerning fifth powers")

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

The Diophantine equation,

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

for either $k=5$ or $6$ is quite well explored, and it has long been known that it has an infinite number of primitive solutions (as points on an elliptic curve, or an infinite family of polynomials,). However,

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

for either $k=7$ or $8$, results seem to be harder to find. Only one solution found back in 2006 is known for $k=8$. Many solutions exist when $k=7$, but it is not known if there are infinite many. Choudhry reduced $(2)$ when it is true for $k=1,3,7$ plus a fourth constraint (re Wolfgang's comment below), namely $\sum\limits^4 x_i = 0$ (and since $k=1$, then also $\sum\limits^4 y_i=0$ ), to a multi-variable cubic equation by passing through the simultaneous equations, in Choudhry's notation,

$$\begin{aligned} X_1X_2X_3\,&=Y_1Y_2Y_3\\ 2(X_1^4 + X_2^4 + X_3^4) - 5(X_1^2 + X_2^2 + X_3^2)^2\, &= 2(Y_1^4 + Y_2^4 + Y_3^4) - 5(Y_1^2 + Y_2^2 + Y_3^2)^2 \end{aligned}\tag3$$

Using that cubic, Choudhry (3) and Wroblewski (21) found a total of $3+21=24$ solutions to $(2)$ valid for $k=1,3,7$. Results can be found in euler.free.fr. Using the $x_i,y_i$, I derived their $X_i, Y_i$. Surprisingly, seven had a simple linear constraint, namely $X_3=2Y_3$. Let,

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

then solutions are,

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

It was hard to find commonalities for the other seventeen solutions.

Questions:

1. Can the simultaneous eqns $(3)$, or the cubic described by Choudhry in the paper, be reduced to an elliptic curve?
2. Why does almost $1/3$ of the $24$ known solutions have $X_3=2Y_3$? Translated into the addends $x_i,y_i$ of $(2)$, this is equivalent to the 5th constraint that $x_1+x_2 = 2y_1+2y_2$. Is there some identity behind it? For example, I found this 7th deg multi-grade,$$1 + 5^k + (3+2y)^k + (3-2y)^k + (-3+3y)^k + (-3-3y)^k = \\(-2+x)^k + (-2-x)^k + (5-y)^k + (5+y)^k$$ for $k = 1,3,5,7,\;$ if $x^2-10y^2 = 9$, guaranteeing an infinite supply with the constraint $x_2 = 5x_1$, though I doubt anything for $(2)$ would be as simple as this.

Note:

However, the fifth degree version,

$$x_1^5+x_2^5+x_3^5 = y_1^5+y_2^5+y_3^5$$

$$x_1+x_2+x_3 = y_1+y_2+y_3=0$$

does have a polynomial identity behind it, found by Choudhry and Wrobleski. (See "A quintic Diophantine equation with applications to two Diophantine systems concerning fifth powers")

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