On the cubic generalization $(a^3+b^3+c^3+d^3)(e^3+f^3+g^3+h^3 ) = v_1^3+v_2^3+v_3^3+v_4^3$ for the Euler four-square

Question

We are familiar with the Euler Four-Square identity,

$$(a^2+b^2+c^2+d^2)(e^2+f^2+g^2+h^2 ) = u_1^2+u_2^2+u_3^2+u_4^2$$

where,

$$u_1 = ae-bf-cg-dh\\ u_2 = af+be+ch-dg\\ u_3 = ag-bh+ce+df\\ u_4 = ah+bg-cf+de$$

or the product of two sums of four squares is itself a sum of four squares.

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Tinkering about, I came across a cubic version,

$$(a^3+b^3+c^3+d^3)(e^3+f^3+g^3+h^3 ) = 6^{-3} (w_1^3+w_2^3+w_3^3+w_4^3)$$

where,

$$w_1= 9 + a^3 + b^3 + c^3 + d^3 + e^3 + f^3 + g^3 + h^3\\ w_2= -9 - a^3 - b^3 - c^3 - d^3 + e^3 + f^3 + g^3 + h^3\\ w_3= 9 - a^3 - b^3 - c^3 - d^3 - e^3 - f^3 - g^3 - h^3\\ w_4= -9 + a^3 + b^3 + c^3 + d^3 - e^3 - f^3 - g^3 - h^3$$

Q: Does the cubic version have any number theoretic implications, like the set of sums of four cubes is closed under multiplication? Or is it just an interesting curiosity?

Answer 1

Summarizing some comments and adding a bit on top:

The identity is a special case of $$24ABC = (\underbrace{A+B+C}_{w_1})^3 + (\underbrace{-A+B-C}_{w_2})^3 + (\underbrace{-A-B+C}_{w_3})^3 + (\underbrace{A-B-C}_{w_4})^3 \tag{1}$$ where setting $C=9$ turns the left-hand side into $6^3AB$. The settings $$\begin{align} A &= a^3 + b^3 + c^3 + d^3 \\ B &= e^3 + f^3 + g^3 + h^3 \end{align}$$ are not necessary for $(1)$ but lead to an interesting specialization.

To make the right-hand side of $(1)$ a symbolic integer multiple of $6^3$, we might want to require that each $w_i$ is divisible by $6$. Straightforward calculations show that this happens if and only if both $A$ and $B$ are divisible by $3$ and $A+B$ is odd. In other words, $$\{A,B\}=\{6m,6n+3\}\quad\text{for some}\quad m,n\in\mathbb{Z}$$

Conversely, for every choice of $m,n\in\mathbb{Z}$ there exist representations of $A$ and $B$ as sums of four cubes, such as $$\begin{align} 6m &= (m+1)^3 + (m-1)^3 + (-m)^3 + (-m)^3 \tag{2} \\ 6n+3 &= n^3 + (-n + 4)^3 + (2n - 5)^3 + (-2n + 4)^3 \tag{3} \end{align}$$ taken from the Alpertron hyperlinked in Dietrich Burde's comment.

Specializing your identity to that scenario gives $$18\,m\,(2n+1) = (2 + m + n)^3 + (-1 - m + n)^3 + (1 - m - n)^3 + (-2 + m - n)^3 \tag{4}$$ which may give much smaller solutions for representations of $18q$ than $(2)$, particularly if $q$ has an odd divisor near $\sqrt{2|q|}$ which we can use for $2n+1$.

Examples: $$\begin{array}{r|rr|rrr} 18q & (2)\quad\text{with} & m & (4)\quad\text{with} & m, & n \\\hline 18 & 4^3 + 2^3 + (-3)^3 + (-3)^3 & 3 & 3^3 + (-2)^3 + 0^3 + (-1)^3 & 1 & 0 \\ 504 & 85^3 + 83^3 + (-84)^3 + (-84)^3 & 84 & 9^3 + (-2)^3 + (-6)^3 + (-1)^3 & 4 & 3 \end{array}$$ The following points may be worth mentioning:

• Replacing $2$ with $0$ in the right-hand side of $(4)$ turns the left-hand side into $6\,m\,(2n-1)$ which is more widely applicable.
• Replacing $2$ with $1$ in the right-hand side of $(4)$ turns the left-hand side into $24mn$, an easy specialization of $(1)$.
• $(1)$ itself, where applicable, might give solutions with even smaller absolute maximum of the numbers involved, particularly if $A,B,C$ can be chosen to be pairwise close.