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I'm struggling with a problem I came around:

Begin by adding $n$ square numbers, no repeats allowed and all bases $\leq$ K, some constant, to reach a sum $s$. For example, $s = 3^2 + 5^2 + 7^2 + 12^2 + 14^2$, where $n = 5$ and $K = 15$.

Is this representation unique? If not, how many sets of length $n$ of square numbers distinctly different from the original set, still following the same conditions described above, exist such that they all sum to $s$?

Order and sign does not matter, e.g. $\{(3)^2, (-5)^2\}$ is considered the same as $\{(5)^2, (3)^2\}$

To generalize this, increase the exponent, $e$ to, say, $=3$. As $e$ increases, how are the other possible sets (if they exist) affected?

(I am a beginner at number theory, so please keep answers as rudimentary as possible. First time posting, so please excuse any formatting issues)

Thanks!

Spaggy
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Mathee
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  • Related post – angryavian Jul 31 '17 at 19:05
  • Not quite sure if this helps, but there are numbers which can be expressed in multiple ways as the sum of powers. – platty Jul 31 '17 at 19:05
  • $9^2+7^2=11^2+3^2$, $1^3+12^3=9^3+10^3$ – fonfonx Jul 31 '17 at 19:19
  • @fonfonx Thank you for the counter-example. Those are cases when n=2. Are there any examples for increased n? (perhaps a link) – Mathee Jul 31 '17 at 19:34
  • @Mathee sorry I don't have a link... I played a bit with square numbers and I found this example, but we can think of plenty of those – fonfonx Jul 31 '17 at 19:51
  • A simple pigeonhole argument (no number theory) shows that if $n > e$, then for sufficiently large $K$ there will be a non-unique representation. And if you allow $K$ to grow arbitrarily large, then you can get any finite number of coincidences that you care to wish for. – Erick Wong Jul 31 '17 at 19:52

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There are several integers that can be expressed as the sum of distinct powers in several ways. It is actually pretty fun to look for them.

For example, for $n=2$,

  • $7^2+9^2=3^2+11^2=130$
  • $1^3+12^3=9^3+10^3=1729$ (the well-known taxicab number of Ramanujan)

For $n=3$, $3^2+4^2+11^2=1^2+8^2+9^2=146$.

For $n=4$, $1^2+2^2+8^2+11^2=4^2+5^2+7^2+10^2=190$.

fonfonx
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Fix any $n$ and $e$, and suppose that $n>e$. Then the number of possible combinations is $\binom{K}{n}$, which is asymptotic to $K^n/n!$, but all combinations are all most $nK^e$ in size.

For large enough $K$, $\binom{K}{n} > nK^e$, so by the pigeonhole principle there will be two combinations that add up to the same value. In fact, as $K \to \infty$, the ratio $\binom{K}{n} / (nK^e)$ is asymptotic to $K^{n-e} / (n+1)! \to \infty$, so for any number $q$, you can find some $K$ such that there are at least $q$ combinations with the same sum.

The case where $n \le e$ will probably require more delicate construction rather than this naive counting argument.

Erick Wong
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