There's a nice identity by Vandermergel. If,
$$a^3+b^3 = c^3+d^3\tag1$$
then,
$$(ac)^3+(bc)^3+(d^2)^3=(ad)^3+(bd)^3+(c^2)^3\tag2$$
Here's one by yours truly. If,
$$a^4+b^4 = c^4+d^4\tag3$$
then,
$$(a^2 + d^2)^2 - (a^2 - d^2)^2 + (2 b c)^2 = (b^2 + c^2)^2 - (b^2 - c^2)^2 + (2 a d)^2\tag{4a}$$ $$(a^2 + d^2)^4 + (a^2 - d^2)^4 + (2 b c)^4 = (b^2 + c^2)^4 + (b^2 - c^2)^4 + (2 a d)^4\tag{4b}$$
Question: Any known $5$th deg identity that will lead from $(5)$ to $(6)$ below,
$$\sum_{i=1}^m u_i^5 = \sum_{i=1}^m v_i^5\tag5$$
$$\sum_{i=1}^n x_i^5 = \sum_{i=1}^n y_i^5\tag6$$
where, like the previous two examples have $m<n$, or $(5)$ has less terms than $(6)$?
P.S. Incidentally, since $(2)$ has $x_1^3+x_2^3+x_3^3=y_1^3+y_2^3+y_3^3$ as well as $x_1x_2x_3=y_1y_2y_3$, then it obeys the high-power relation,
$$3(x_1^3+x_2^3+x_3^3)(x_1^6+x_2^6+x_3^6-y_1^6-y_2^6-y_3^6)=2(x_1^9+x_2^9+x_3^9-y_1^9-y_2^9-y_3^9)$$
though I haven't explored the properties of $(4)$ yet.
