(*This is an addendum to my answer*.)

**I. Identities.** For those curious on further generalizations of Ramanujan's 6-10-8, while *Theorem 3* used $2^3$ terms, one with higher powers uses $2^5$ terms. Define,

$$\small P_n = \sum^{16}\, (\mu(a_1\pm a_2\pm a_3\pm a_4\pm a_5))^n - \sum^{16}\, (\mu(b_1\pm b_2\pm b_3\pm b_4\pm b_5))^n\tag1$$

where $\mu = \pm1$ and is the product of the interior signs. If *four* conditions are now satisfied,

$$\small\prod^5 a_i = \prod^5 b_i$$

$$\small \sum^5 a_i^k = \sum^5 b_i^k$$

for $k = 2,4,6$, then,

$$P_1 = P_2 = P_3 = P_4 = P_5 = P_6 = P_7 = P_9 = P_{11} = 0$$

and,

$$957\,P_8 P_{15} = 1547\,P_{10} P_{13}$$

**II. Families.** I found one can solve the four conditions in two ways: the first in terms of *quadratic forms*, and the second as an *elliptic curve*.

**Family 1:** If $x^2+21y^2 = z^2$, then,

$$a_i = x + 6 y,\; x + 5 y - z,\; x + 5 y + z,\; \tfrac{1}{2}(-5x+2y),\; \tfrac{3}{2}(x-6y)$$

$$b_i = x - 6 y,\; -x + 5 y - z,\; -x + 5 y + z,\; \tfrac{1}{2}(5x+2y),\; \tfrac{3}{2}(x+6y)$$

**Family 2:** If $-4a^2+5b^2=4c^2,\;\; 25a^2+24b^2=d^2$, then,

$$a_i = 4 a - 4 b,\; 3 b - 2 c,\; 3 b + 2 c,\; 6 a,\; 4 a + 4 b$$

$$b_i = b - 2 c,\; b + 2 c,\; 4 a,\; -a + d,\; a + d$$

*Note:* For the second family, eight terms will cancel out in $(1)$, so it really involves only $2^5-8=24$ terms. An initial solution is $a,b,c,d = 29, 26, 2, 193$ and, using an elliptic curve, one can get an infinite more. Explicitly,

$$\small( 281, -207, -199, -107, \color{blue}{-61}, 125, \color{blue}{33}, -13, \color{blue}{25}, -21, \color{blue}{-113}, 49, 95, 187, 195, -269)^n=\\
\small(277, -247, -161, \color{blue}{-113}, -55, 131, 83, \color{blue}{25}, -3, \color{blue}{-61}, -109, \color{blue}{33}, 91, 139, 225, -255)^n$$

for $n = 1,2,3,4,5,7,9,11$. (The second family was found in 2013 with the help of *Roger Glendenning* who provided numerical solutions. After some trial and error, I found the form above.)