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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).

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    From [Pythagorean Quadruple](http://mathworld.wolfram.com/PythagoreanQuadruple.html): *Oliverio gives the following generalization of this result. Let $S=(a_1,...,a_{n-2})$, where $a_i$ are integers, and let $T$ be the number of odd integers in $S$. Then iff $T≢2 \mod 4$, there exist integers $a_{n-1}$ and $a_n$ such that $ a_1^2+a_2^2+...+a_{n-1}^2=a_n^2. $* – 2012-04-02

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Here's an example with odd numbers: $1^3+9^3+15^3+23^3=3^3+5^3+19^3+21^3$

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    @draks, the depths of my ignorance are limitless. The more I can do, the more I see there is that I can't do. – 2012-04-03
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If you look at the small summary about "taxicab" numbers, you will see that there are a fair number of $(s,t)$, $(u,v)$ such that $s^3+t^3=u^3+v^3$ and $s$ and $t$ are both odd, while $u$ and $v$ are both even.

By looking at tables of Hardy-Ramanujan numbers, we can produce examples that have more cubes. For $k=4$, all we need to do is to find in the tables two numbers $A$ and $B$ which can each be expressed as the sum of two odd cubes and also of two even cubes. Then the sum $A+B$ can be expressed as the sum of four odd cubes, and also of four even cubes.

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    There are infinitely many examples for every $r$. One way to interpret your question is, given $k,r,x$, how many nontrivial solutions are there to $\sum^ka_i^r=\sum^kb_i^r$ with the sum not exceeding $x$ (or with the individual $a_i$ and $b_i$ not exceeding $x$). This is the kind of problem traditionally handled by the Hardy-Littlewood Circle Method from Analytic Number Theory. Some cases are solved (asymptotically), some have conjectured solutions, some are wide open. No one knows whether there's a nontrivial solution to $a^5+b^5=c^5+d^5$. – 2012-04-02