Like my previous question, I'll pose this one too with an array.
$1^r, 3^r, 5^r, 7^r, 9^r$ (all odd $r$th powers)
That's array 1. And array 2;
$2^r, 4^r, 6^r, 8^r, 10^r$ (all even $r$th powers)
Let's take the sum of random $k$ integers from each array, where $r>2$ (to eliminate Pythagorean triplets), so our $r=3, k=3$, and from Array 1, we have $(1^3 + 5^3 + 7^3)=469$ and for Array 2, $(6^3 + 10^3 + 2^3)=1224$
Now let me get to the point. I want to know, is it possible for the sum of $k$ random odd/even $r$th powers from an array of consecutive odd/even $r$th powers respectively, to be partitioned in any other way than the $r$th powers that constituted them?
I guess the answer for this must be $1$ (the initial sum of $r$th powers)...
PS I've taken separate arrays to avoid the trouble of Hardy-Ramanujan-like numbers (as in cubes) and equivalence to other powers (like $3^3 + 4^3 + 5^3=6^3$) from occuring in the sum. The effort is to keep the possible partitions as close as possible to $1$ (initial composition).