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Edward Waring, asks whether for every natural number $n$ there exists an associated positive integer s such that every natural number is the sum of at most $s$ $k$th powers of natural numbers (for example, every number is the sum of at most 4 squares, or 9 cubes, or 19 fourth powers, etc.) from here.

I ask: Are there unique solutions for $n=\sum_{j=1}^{g(k)} a_j^k$, with $a_u\neq a_v \geq0 \; , \; \forall u\neq v$?

$g(k)$ being the minimum number $s$ of $k$th powers needed to represent all integers. $14$ would be an example of a unique decomposition (into $0^2+1^2+2^2+3^2$).

Non-unique decompositions for $k=2$ can be constructed by letting $ n=(a_0+x)^2+\sum_{j=1}^3 a_j^2$ and $a_0=x+\sum_{j=1}^3 a_j$. Robert gave a formula for cubes below.

So another question is: How can one test that a certain $n$ has a unique solution, or even better how can I calculate the number of respresentations? As Gerry points out in his comment, a prime $p=4k+1\;$ has a unique representation $p=a^2+b^2$ with $0Thue's Lemma).

Partial answer for $k=2$ (from here)

The sequence of positive integers whose representation as a sum of four squares is unique is:

1, 2, 3, 5, 6, 7, 8, 11, 14, 15, 23, 24, 32, 56, 96, 128, 224... (sequence A006431 in OEIS).

These integers consist of the seven odd numbers $1, 3, 5, 7, 11, 15, 23$ and all numbers of the form $2 × 4^k, 6 × 4^k$ or $14 × 4^k$.

Partial, therefore, because $23=1^2+2^2+2\times 3^2$ (answer, because $14\times 4^k$ is unique).


Another way to look at it, uses ${g(k)}$-dimensional vector spaces $V$ over $\mathbb N_0^+$ that are provided with an $k$-norm $||a||_k=\left( \sum_{j=1}^{g(k)} a_j^k \right)^{1/k}$ and the additional conditions that $a_u\neq a_v \geq0 \; , \; \forall u\neq v$. Now there is a set of length-preserving operations $U_p$, that transforms vectors, from a defined range $\mathcal R (U_p)$. An example operation was shown above. In this framework the question sounds like:

Which vectors of $V$ lie outside the union of all $\mathcal R (U_p)$?

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    Note that $\sum_{j=1}^ka_j^j$ is an abbreviation for $a_1+a_2^2+a_3^3+\cdots+a_k^k$; $k=2$ would be $n=a_1+a_2^2$, which has nothing to do with Lagrange. What do you really want? Please edit your question accordingly. A prime $p=4k+1$ has a unique representation $p=a^2+b^2$ with $0\lt a\lt b$. There are formulas for the number of representations as a sum of 4 squares, readily available in texts and on the web.2012-01-29
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    @GerryMyerson: (i) There are formulas, e.g. [here](http://en.wikipedia.org/wiki/Jacobi%27s_four-square_theorem) i found something by Jacobi, but he allows values to appear multiply, see example there. (ii) What does it like for cubes,...?2012-01-30
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    There is a (relatively simple to prove) theorem due to Sprague (from 1940s) stating that all the integers larger than 128 can be written as sums of unequal squares. 128 cannot be written as a sum of unequal squares, so that variant of the question is fully settled. Similar results exist for cubes (and higher). I don't know about Waring type of bounds on the number of terms, though. They would probably work for large integers only. Not really your variant, I know. May be a useful buzzword when looking for more information?2012-01-30
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    @JyrkiLahtonen: Is there an online reference to that theorem?2012-01-30
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    I don't know. The idea is simple. You show by brute force checking that you can always pick a subset of $1^2,2^2,3^2,\ldots,10^2$ such that it has any desired sum $n$ in the range $129\le n\le192$. Then the complements of those sums give you similar representations for numbers $m$ in the range $385-192=193\le m\le 385-129=256.$ ($\sum_{k=1}^{10}k^2=385$) This range $[129,256]$ is longer than $11^2$, so adding that term to the mix proves it for all the integers up to $256+11^2=377$. Then add $12^2$ et cetera.2012-01-30
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    @JyrkiLahtonen: But this doesn't take into account that I want to use just $g(k)$ elements of the set.2012-01-30
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    No, it doesn't. That's what I meant, when I said that *I don't know about bounds on the number of terms*, and that *this is not exactly your variant of the question*. You most certainly need more than $g(2)=4$ squares to get all the (large enough) numbers represented as sums of distinct squares, and some cannot be written in that way at all - $2$ being the most obvious example. If you are only interested in uniqueness of the solution, then this is probably not very useful to you.2012-01-30
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    An example for cubes of the kind of identity you're considering is $(a_1 + 2 x)^3 + \sum_{j=2}^9 a_j^3 = a_1^3 + \sum_{j=2}^9 (a_j + x)^3$ where $x = {\frac {-2\,{a_{{1}}}^{2}+{a_{{2}}}^{2}+{a_{{3}}}^{2}+{a_{{4}}}^ {2}+{a_{{5}}}^{2}+{a_{{6}}}^{2}+{a_{{7}}}^{2}+{a_{{8}}}^{2}+{a_{{9}}}^ {2}}{4\,a_{{1}}-a_{{2}}-a_{{3}}-a_{{4}}-a_{{5}}-a_{{6}}-a_{{7}}-a_{{8} }-a_{{9}}}} $2012-01-31
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    @RobertIsrael: Nice! Do you think there is a way to describe all operations?2012-01-31
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    F Halter-Koch, Darstellung naturlicher Zahlen als Summe von Quadraten, Acta Arith 42 (1982) 11-20 proves that if $n\gt157$ is odd then it is a sum of four distinct non-zero coprime squares. The result is used in Bateman, Hildebrand, Purdy, Sums of distinct squares, Acta Arith 67 (1994) 349-380, which is available at http://matwbn.icm.edu.pl/ksiazki/aa/aa67/aa6745.pdf2012-01-31
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    @GerryMyerson Great reference!2012-01-31
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    Also of possible interest is http://oeis.org/A001661 "Largest number not the sum of distinct n-th powers."2012-01-31
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    According to Guy, Unsolved Problems In Number Theory, Problem C20, the Halter-Koch paper also proves that every $n\gt412$ and not a multiple of 8 is a sum of four distinct non-zero squares.2012-01-31

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