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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).

Mach9
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  • Maybe you are interested in this... – draks ... Apr 02 '12 at 11:39
  • From Pythagorean Quadruple: 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 $$ a_1^2+a_2^2+...+a_{n-1}^2=a_n^2. $$ – draks ... Apr 02 '12 at 11:47

2 Answers2

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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.

André Nicolas
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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? – Mach9 Apr 02 '12 at 12:09
  • My guess is, it might be a black-or-white solution to this question-mostly, no, IMHO. – Mach9 Apr 02 '12 at 12:21
  • @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 $k$? – André Nicolas Apr 02 '12 at 12:25
  • 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? – Gerry Myerson Apr 02 '12 at 12:35
  • @André Nicolas: Yes. – Mach9 Apr 02 '12 at 13:11
  • @Gerry Myerson: Yes, sort of... – Mach9 Apr 02 '12 at 13:12
  • @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? – Mach9 Apr 02 '12 at 14:24
  • And yes, what about higher powers like $r=4, 5, 6...$ (anything so far $r>2$)? – Mach9 Apr 02 '12 at 14:25
  • 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 $a^5+b^5=c^5+d^5$. – Gerry Myerson Apr 02 '12 at 23:09
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Here's an example with odd numbers: $$1^3+9^3+15^3+23^3=3^3+5^3+19^3+21^3$$

Gerry Myerson
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  • +1 nice. How did you get that? – draks ... Apr 02 '12 at 12:54
  • 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. – Gerry Myerson Apr 02 '12 at 13:09
  • The Prouhet–Tarry–Escott problem? – draks ... Apr 02 '12 at 13:12
  • @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? – Mach9 Apr 02 '12 at 14:27
  • And yes, what if the number has to be a cube corresponding to an element in the array? – Mach9 Apr 02 '12 at 14:28
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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. – Gerry Myerson Apr 02 '12 at 23:03
  • @GerryMyerson nice, is there anything you can't do? – draks ... Apr 03 '12 at 06:17
  • @draks, the depths of my ignorance are limitless. The more I can do, the more I see there is that I can't do. – Gerry Myerson Apr 03 '12 at 07:45