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We have the polynomial identity

$360\ (2X + 1) = (X+1)^{5} + (X+4)^{5} + (-X -3)^{5} + (-X -3)^{5} - (X-2)^{5} - (X-2)^{5} + (X-3)^{5} + X^{5}.$

Thus, every rational integer that is the product of $360$ by an odd rational integer is the sum of $8$ fifth powers of rational integers.

If $x$ is a rational integer prime to $6$, then $x$ is of the form $y^{5} + 360\ (2t + 1)$, with $y$ and $t$ rational integers. Thus, every rational integer prime to $6$ is a sum of $9$ fifth powers of rational integers. Thus every odd rational integer $n$ is a sum of $10$ fifth powers of rational integers. (If $n$ is prime to $3$, it results a fortiori from the preceding fact. If $n$ is divisible by $3$, $n - 2^{5}$ is a sum of $9$ fifth powers by the preceding fact.) Also, every even rational integer $n$ is the sum of $10$ fifth powers of rational integers. (Indeed, $n - 1^{5}$ or $n - 3^{5}$ is prime to $6$ and is thus a sum of $9$ fifth powers of rational integers by a preceding fact.)

Thus every rational integer is a sum of $10$ fifth powers of rational integers.

My question is : can this result be improved, in other words : is it proven that every rational integer is the sum of $9$ fifth powers of rational integers ? Thanks in advance for the answers.

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  • $\begingroup$ The Wolfram Mathworld page on Warings problem states: "If the negatives of powers are permitted in addition to the powers themselves, the largest number of nth powers needed to represent an arbitrary integer are denoted eg(n) and EG(n) (Wright 1934, Hunter 1941, Gardner 1986). In general, these values are much harder to calculate than are g(n) and G(n)." The references (which may be useful for you are): $\endgroup$ – Vincent Mar 15 '17 at 9:05
  • $\begingroup$ Wright, E. M. "An Easier Waring's Problem." J. London Math. Soc. 9, 267-272, 1934. Hunter, W. "The Representation of Numbers by Sums of Fourth Powers." J. London Math. Soc. 16, 177-179, 1941. Gardner, M. "Waring's Problems." Ch. 18 in Knotted Doughnuts and Other Mathematical Entertainments. New York: W. H. Freeman, pp. 222-231, 1986. $\endgroup$ – Vincent Mar 15 '17 at 9:05
  • $\begingroup$ (Ok, apparently it is impossible to insert linebreaks into MSE-comments. I hope the previous comment is still somewhat readable) $\endgroup$ – Vincent Mar 15 '17 at 9:07
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The question of how many $k$-th powers are needed to represent every number is called Waring's problem. It is clear that Waring's problem for fields is in general easier than for rings. Nevertheless it is still difficult to find explicit numbers $g(k)$ for the minimal number of $k$-th powers needed. For $\mathbb{Q}$ we know that $g(\mathbb{Q},2)=4$, $g(\mathbb{Q},3)=3$, $g(\mathbb{Q},4)=15$. Unfortunately it seems that $g(\mathbb{Q},5)$ is not known, see here, end of page $4$, and beginning of page $5$. However, there is the following claim:

Any rational number is the sum of six rational fifth powers - see here.

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