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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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    Maybe you are interested in [this](http://math.stackexchange.com/q/103432/19341)...2012-04-02
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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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    +1 nice. How did you get that?2012-04-02
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    I know a way to get numbers satisfying $a^i+b^i+c^i+d^i=w^i+x^i+y^i+z^i$ simultaneously for $i=1,2,3$. These are called "multigrade equations", I'm sure there's info about them on the web. If you replace each base $n$ appearing in the equation with $2n+1$ it's still a multigrade, and now all the numbers are odd.2012-04-02
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    The [Prouhet–Tarry–Escott problem](http://en.wikipedia.org/wiki/Prouhet%E2%80%93Tarry%E2%80%93Escott_problem)?2012-04-02
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    @Gerry Myerson: Right. But then, what if I were to say all the cubes are to be even? Would I have any other representations (for the sum of random $k$ cubes) as other $k$ cubes?2012-04-02
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    And yes, what if the number _has_ to be a cube corresponding to an element in the array?2012-04-02
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    @draks, yes. Mach9, I already made the cubes even in my comment on Andre's answer. I can make them all end in 6, if you want: $16^3+96^3+156^3+236^3=36^3+56^3+196^3+216^3$. That's the beauty of multigrades. "what if the number has to be a cube corresponding to an element in the array" - what if *what* number has to be a cube? *what* array? I don't understand what you are asking.2012-04-02
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    @GerryMyerson nice, is there anything you can't do?2012-04-03
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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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    Uh, I mean to ask, _is it possible for, as in the example, the sum of cubes of $k$ random even/odd numbers to be the sum of other $k$ (even/odd, respectively) cubes?_2012-04-02
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    My guess is, it _might_ be a black-or-white solution to this question-mostly, no, IMHO.2012-04-02
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    @Mach9: Certainly the phenomenon is rare. But you do not seem to be asking for a measure of how rare. The link I gave has several examples of two odd positive cubes which have the same sum as two even positive cubes. so it it gives some examples for your case $r=3$, $k=2$. Are you asking about other $k$?2012-04-02
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    If you take an example of two odd cubes equal two even cubes, and multiply through by 8, you'll have two even cubes adding up to the same number as two other even cubes. Mach9, isn't that what you want?2012-04-02
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    @André Nicolas: Yes.2012-04-02
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    @Gerry Myerson: Yes, sort of...2012-04-02
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    @André Nicolas: Now that I find you can have many representations for the same number of cubes, just how many would there be, and how can I calculate it?2012-04-02
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    And yes, what about higher powers like $r=4, 5, 6...$ (anything so far $r>2$)?2012-04-02
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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