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

I've asked the same question at the Math Stack Exchange site, but I didn't have any luck there. So I'm posting the same question here.

Denote by $r_{s,k}(x)$, the number of ways in which $x$ can be expressed as the sum of $k$ $s^{\text{th}}$ 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 natural $a$, $r_{s,k}(a^s)$ 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 if it's hopeless to do so. :)

Many thanks in advance!

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It has been 19 excruciating hours. :) Oh well, I agree anyways. – Koundinya Vajjha May 11 '11 at 8:53
@Theo: if it matters, the OP has waited more hours than his age in years. :) – Pete L. Clark May 11 '11 at 11:01
@Theo: I've added the link. :) – Koundinya Vajjha May 11 '11 at 11:39

The negation of the statement $$\forall s\in \mathbb{N}\; \exists\: k \; \: \text{such that} \;r_{s,k}(x)\geq 1\; \forall\; x \in \mathbb{N}$$ is not the statement $$\forall s\in \mathbb{N}\; \nexists\: k \; \: \text{such that} \;r_{s,k}(x)\geq 1\; \forall\; x \in \mathbb{N}$$ but the statement $$\exists s\in \mathbb{N}\; \forall\: k \; \exists x \in \mathbb{N}\: \text{such that} \;r_{s,k}(x)=0$$ In other words: if the Hilbert-Waring theorem is false, then there is some exponent $s$ such that for any number of summands $k$ there exists a natural number $x$ which is not the sum of $k$ $s$-th powers. Your confusion comes from the fact that you made a basic error in logic.
EDIT 2: I should also add that in the theory of the Waring problem one usually denotes by $k$ the exponent and by $s$ the number of summands, not the other way.
Thank you very much for your answer GH. :) I wasn't expecting a trivial proof at all. I just was toying with this function and I just thought I'd have a go at the Hilbert-Waring theorem. And I do know that the $s$ and $k$ should be interchanged, but I've gotten used it. Sorry for that. And I'll go through the answer tomorrow. It is pretty late now. Thank you for your answer again. – Koundinya Vajjha May 11 '11 at 15:34