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Denote by $r_{s,k}(x)$, the number of ways in which $x$ can be expressed as the sum of $k$ $s^{th}$ powers of integers. Now, the Hilbert-Waring theorem is equivalent to the following. $\forall s\in \mathbb{N}\; \exists\: k \; \: \text{such that} \;r_{s,k}(x)\geq 1\; \forall\; x \in \mathbb{N}$  Now, call me optimistic, but I've been trying to prove this fact via contradiction. So, assuming the contradiction gives us the following. $\forall s\in \mathbb{N}\; \nexists\: k \; \: \text{such that} \;r_{s,k}(x)\geq 1\; \forall\; x \in \mathbb{N}$  Now, if this is true, then it means that there is at least one $s$ for which there exists a corresponding $k$, such that $r_{s,k}(x)\geq 1$. So, for all the other integers $s$, the opposite must hold.

I have come this far till now. If we can show somehow that from the above arguments it must follow that $r_{s,k}(x)=0$ the proof will follow, since for any integer $a$, $r_{s,k}(a^s)$ is always greater than or equal to one.

I'd be grateful to anyone who pitches in any ideas on how to approach this problem, or of it's hopeless to do so. :)

Many thanks in advance!

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    [Crossposted to MO](http://mathoverflow.net/questions/64578/the-hilbert-waring-theorem-using-the-sum-of-squares-function)2011-05-11

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Since this is unanswered:

Your confusion comes from incorrectly negating the statement.