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Apr 13, 2017 at 12:19 history edited CommunityBot
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Oct 1, 2014 at 16:24 comment added Guy German is indeed the problem...
Oct 1, 2014 at 14:37 comment added Emil Jeřábek Fricker’s book is easily found on the internet (with a choice between Springer and pirates), and does indeed contain a proof of the result, based on Jacobi's four-square theorem. It only holds in the form you stated it for $d\ge5$, whereas for $d=4$ one has $V_4R^4+O(R^2\log R)$. The result with the logs is due to Landau, the stronger form for $d\ge5$ apparently to Walfisz. There are several other references in the book, all German, if that’s the problem.
Oct 1, 2014 at 13:43 comment added Guy @Joonas, I think you need $d \geq 4$. Anyway I'm interested in the case where $d$ is sufficiently large. Chamizo and Iwniec make the claim for $\alpha=d-2$ in the second paragraph in the paper I linked to. The say there is an elementary proof for it and cite the book by Fricker. This is the proof I'm interested in.
Oct 1, 2014 at 12:28 comment added Emil Jeřábek Chamizo and Iwaniec only state it as $V_dR^d+R^{d-2+\epsilon}$ for any $\epsilon>0$, and that it is proved for $d\ge4$. They also mention that the optimal exponents for $d=2$ and $d=3$ are conjectured to be $1/2+\epsilon$ and $1+\epsilon$, respectively, that the best upper bound proved for $d=2$ (at the time) is $46/73+\epsilon$, and the main result of the paper is a $29/22+\epsilon$ upper bound for $d=3$.
Oct 1, 2014 at 12:08 comment added Joonas Ilmavirta If $d\leq2$, is $\alpha$ still $d-2$?
Oct 1, 2014 at 12:07 review First posts
Oct 1, 2014 at 12:07
Oct 1, 2014 at 12:05 history asked Guy CC BY-SA 3.0