# Hilbert-Waring theorem using the sum-of-squares function

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. :)

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Your $\mathbb{Z}$s need to be $\mathbb{N}$, or $s=2, x=-1$ is an easy counterexample. Then when you interpret the assumption, it means there is at least one $s$ where no specific $k$ is sufficient for all $x$. –  Ross Millikan May 10 '11 at 13:14
Oh. Yeah. Will edit that. Thanks. –  kodyv May 10 '11 at 13:57
Crossposted to MO –  t.b. May 11 '11 at 8:42