Using Rachinsky quintets and others to find $8\text{th}$ powers $x_1^8+x_2^8+x_3^8+x_4^8+x_5^8=y_1^8+y_2^8+y_3^8+y_4^8+y_5^8\,$?

Question

I. Rachinsky quintets

In this previous post, it was shown that special Pythagorean quadruples can lead to $6$th powers. We go higher and use special Rachinsky quintets that lead to $8$th powers. In this MO post, there is this nice Russian painting,

In the blackboard, teacher Sergei Rachinsky has written

$$\frac{10^2 + 11^2 + 12^2 + 13^2 + 14^2}{365}$$

One of the solutions is based on the equality $10^2 + 11^2 + 12^2 = 13^2 + 14^2$. The MO post labels these as Rachinsky quintets $(a,b,c,d,e)$ and as satisfying,

$$a^2+b^2+c^2=d^2+e^2$$

---

II. Letac-Sinha Identity

Using a pair of Rachinsky quintets,

$$(a + c)^2 + (a - c)^2 + (4a)^2 = (3a + c)^2 + (3a - c)^2$$ $$(b + d)^2 + (b - d)^2 + (4b)^2 = (3b + d)^2 + (3b - d)^2$$

already valid for any $(a,b,c,d)$. But if we combine them, it surprisingly becomes valid for multiple powers including the $8\text{th}$, hence,

$$(a + c)^k + (a - c)^k + (3b + d)^k + (3b - d)^k + (4a)^ k = \\(3a + c)^k + (3a - c)^k + (b + d)^k + (b - d)^k + (4b)^k$$

true for $k = (1,2,4,6,8)$ with the condition that $(a,b,c,d)$ are chosen so,

$$a^2+12b^2=c^2\\12a^2+b^2=d^2$$

Given an appropriate rational point, one can then transform this to an elliptic curve hence there are infinitely many such primitive $(a,b,c,d)$.

But one should avoid the ratio $a/b=1/2$ and $a/b=2$ as it just leads to trivial solutions. The smallest non-trivial is $(a,b) = (218, 11869)$ hence, after removing common factors, we get the Rachinsky quintet,

$$\color{blue}{20667^2 + (-20449)^2 + 436^2 = 20885^2 + (-20231)^2}$$

which, together with its partner, yields,

$$\color{blue}{20667^8} + \color{blue}{(-20449)^8} + 23750^8 + 11857^8 + \color{blue}{436^8} = \\ \color{blue}{20885^8} + \color{blue}{(-20231)^8} + 11881^8 + (-12)^8 + 23738^8$$

and infinitely more primitive solutions. As expected, the example is multigrade for $k=(1,2,4,6,8)$. Interesting that $12$ is also a term.

Update: Courtesy of Tomita,

$$(a,b) = (218, 11869)\\ (a,b) = (94940348411, 220873077098)$$

so the numbers get large fast, though there are infinitely many of them.

---

III. Question

If one looks at this old Mathworld list of small solutions to $(8,5,5)$, absolutely none are multi-grade even for just $k=(4,8)$. Then in 2000, Borwein, Lisonek, and Percival found by computer search,

$$366^k+ 103^k+ 452^k+ 189^k+ (-515)^k =\\ 508^k+ 245^k+ (-18)^k+ 331^k+ (-471)^k$$

which again is multi-grade for the complete set $k=(1,2,4,6,8)$.

I've arranged the terms in this manner since, like the Letac-Sinha identity, there are linear relations between them. Labeled as $(x_i, y_i)$, then it obeys,

$$x_1-x_2 = y_1-y_2\\ x_1-x_4 = y_1-y_4\\ x_2-x_4 = y_2-y_4\\ x_3-x_4 = y_2-y_3$$

The fact that the terms are relatively small, obey such tidy relationships, and valid for five exponents $k$ (not just one $k$) suggests there is a reason.

Q: Does the 2000 multi-grade equality in fact also belong to an infinite family? (I've been trying to crack this for years.)

Answer 1

As I was reminded by Tomita, the 2000 equality by Borwein et al does belong to a family (though it may belong to several with different constraints). But unlike the Letac-Sinha identity which needs only two quadratics to be made squares, the family I found needs four quadratics. Back in 2010, Wrobleski suggested the system,

$$[x_1,\, x_2,\, x_3,\, x_4,\, x_5]^k = [y_1,\, y_2,\, y_3,\, y_4,\, y_5]^k$$

$$x_1-x_2 = x_3-x_4 = y_1-y_2 = y_2-y_3$$

for $k = (2,4,6,8)$, so an under-determined system of 7 equations in 10 unknowns. The form I used has six variables $(a,b,c,d,e,f)$,

$$[b-c-d, b+c-d, -b-c-d, -b+c-d,\, f]^k = [a+b-2c, a+b, a+b+2c, a-b,\, e]^k$$

which, by systematically eliminating $(d,e,f)$ using resultants leads (after factoring) to a single equation in 10-7 = 3 variables $(a,b,c)$, namely,

$$5a^2-10ab+8b^2 = 32c^2$$

whose low degree was a welcome surprise. However, recovering the other variables, the full set of conditions were four quadratics,

\begin{align} 5a^2-10ab+8b^2 &= 32c^2\\ 29a^2+6ab+72b^2 &= 32d^2\\ 9a^2-6ab+36b^2 &= 4e^2\\ 13a^2+2ab+4b^2 &= 4f^2\end{align}

Thus, one should choose correct $(a,b)$ for integer $(c,d,e,f)$. By using an elliptic curve, there are infinitely many integer $(a,b)$ such that $(c,d)$ are also integers, though the ratios $a/b = (2/3,2,6)$ should be avoided. An example is $(a,b) = (42,37)$ which yields,

$$[-45, -22, -119, -96, \sqrt{7879}]^k = [56, 79, 102, 5, \sqrt{13959}]^k$$

for $k = (2,4,6,8)$. But $(a,b) = (-288,43)$ makes all $(c,d,e,f)$ as integers, yielding the Borwein equality,

$$[366, 103, 452, 189, -515]^k = [508, 245, -18, 331, -471]^k$$

This is the only primitive $(a,b)$ known to do so. To make all the $(x_n,\,y_n)$ as integers, one has to choose a different set of 7 equations that may lead to only two quadratics to be made squares, but no one has found that set yet.