Diophantine Equations and Inequalities in Algebraic Number by Yuan Wang

By Yuan Wang

The circle process has its genesis in a paper of Hardy and Ramanujan (see [Hardy 1])in 1918concernedwiththepartitionfunction andtheproblemofrep resenting numbers as sums ofsquares. Later, in a sequence of papers starting in 1920entitled "some difficulties of'partitio numerorum''', Hardy and Littlewood (see [Hardy 1]) created and constructed systematically a brand new analytic approach, the circle approach in additive quantity concept. the main well-known difficulties in advert ditive quantity thought, particularly Waring's challenge and Goldbach's challenge, are taken care of of their papers. The circle process can be known as the Hardy-Littlewood technique. Waring's challenge can be defined as follows: for each integer okay 2 2, there's a quantity s= s( okay) such that each optimistic integer N is representable as (1) the place Xi arenon-negative integers. This statement wasfirst proved by way of Hilbert [1] in 1909. utilizing their strong circle strategy, Hardy and Littlewood received a deeper consequence on Waring's challenge. They confirmed an asymptotic formulation for rs(N), the variety of representations of N within the shape (1), specifically okay 1 only if eight 2 (k - 2)2 - +5. the following

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Since J3 and J1 + J2 are not empty we have E 1 ~ ATon - 1 ~ AB n - 1. 28). 29) 2) Estimation of 52. L) E 42 Chapter 3. Weyl's Sum 82 L ~ {9, }~{o} First we assume that Wo Since la(t) I W(g)min--,· tEJa Tt Igt > 1. Then Igil ~ Gi, we have . la(t) 1 " , , ' ~ la(g) I mm-- ~ LJ LJ LJ - - , a {9,}~{O} tEJ Ttlgtl gOa (g) 9,=1 Tgiggi " LJ where ~(g). means a sum taken over gt (t E J3 , t -::f q). Therefore the above expreSSlOn 1S and consequently la(t) I 8 2 ~ Wo IT--logh tEJa Tt IT ITIT(j) 1-1 log h ~ hii En-ii IT Tt ITIT(j) 1- 1 log h.

7. 7]) we assume in the following sections that n > l. 2. Suppose that k = 2. 2) does not exceed the number of solutions of the equation Set 1 ~ i ~ 4. Then the number concerned does not exceed the number R(O(T» of solutions of 6(1 + 6(3 = 6(2 + e4(4, ei, (i E M(O(T», 1 ~ i ~ 4. 1 we obtain R(O(T» = O(T6n), and the theorem follows. Suppose now that k ~ 3 and that the theorem holds for k - 1 instead of k. \+6EM'(T) L 1+ AEM'(T) where L AEM'(T) ' 8 =f O. xeM'(T) I E(-Te)dx. L where E* denotes a sum over all integers 81, ...

6) for some index j + rz, and we must prove that ~ rl Since q-l is integral, we have N( q) ~ 1. 8) in the case j > rl. 6). 5) if the pair a, ~ were in 6. 4) are not all satisfied. 8), 0< IIKall ~ h, and so we have IK(i)lla(i)~(i) h la(j)~U) ,8(i) I< i, i =I j or i =I j,j + rz, - ,8U)1 ~ IKU)I- 1 ~ D- 1 / Z • The lemma is proved. 1. 9) = rl + rz. Let 11, ... " r be a set of real numbers satisfying L Ii + 2 L 1m = O. pter 3. Weyl's Sum 1 :5 i :5 r, where C2 = c2(K). Proof. We know by Dirichlet's unit theorem that there always exists a set of fundamental units O'}, ••• , O'r-l for the group of totally nonnegative units of K (see, for example, [Heeke 1, §35J.

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