9 Homogeneous equations and Birch's theorem

9 Homogeneous equations and Birch's theorem
9.1 Introduction Let F(x1,. . . , xs) be a homogeneous form of degree k ^ 2 with integer coefficients. A natural question is to ask whether the equation F(xl9...,xs) = 0 (9.1) has a non-trivial solution, i.e. a solution in integers Xj not all zero. Obviously when k is even the equation may only have the trivial solution. However, when k is odd there is more hope. Lewis (1957) building on earlier work of Brauer (1945) showed that if s is sufficiently large, then any cubic form in s variables with integer coefficients has a non-trivial zero. Shortly afterwards this was extended by Birch (1957) to forms of arbitrary odd degree. Indeed, Birch proved somewhat more than this. The object here is to give an account of Birch's theorem. For references to later work on this and related topics the interested reader should see Davenport's collected works (Davenport, 1977). The proof of Birch's theorem rests on a special case, namely on the solubility of the additive homogeneous equation ^+... + ^ = 0, (9.2) and this can be treated by an application of the Hardy-Littlewood method. 9.2 Additive homogeneous equations The methods of Chapters 2, 4 and 5 are readily adapted to give the following theorem, and so the proof is only given in outline. Theorem 9.1 Let k > 2 and s0 be as in Theorem 5.4, and suppose that s > min(s0, 2k + 1) and s > 4/c2 — k + 1. Suppose further that when k
148 Homogeneous equations and Birch's theorem is even not all of the integers c1, . . . , cs are of the same sign. Then the equation (9.2) has a non-trivial solution in integers x,, . . . , xs. Throughout this section, implicit constants may depend on c c *-" 1J • • • J ^s" If there is ay' such that Cj = 0, then the conclusion is trivial. Thus it may be assumed that, for every j, Cj =£ 0. Also, when k is odd, it can be assumed (if necessary by replacing x1 by — xj that not all the cj are of the same sign. Let R(N) denote the number of solutions of (9.2) with 1 ^ Xj^ N. Then the methods developed in Chapters 2, 4 and 5 give R(N)=&J(N) + 0(Ns-k-d) where OC 3 = n T(p), T(p) = £ S(ph), p /i = 0 q s s($ p>C and that J(N)>Ns~k. Now it suffices to show that T(p) > 0 and, again, this will follow if it is shown that MF{q\ the number of solutions of F(x1? . . . , xs) = cxx* + . . . + csxks = 0(mod q) with 1 < Xj ^ q, satisfies, for t sufficiently large, MF(Pt)>C(p)PHs~1) (9.3) for some positive number C(p) depending only on c1?. . ., cs and p. In order to treat MF it is necessary to transform the variables so as
Additive homogeneous equations 149 to obtain a new form H in which an appreciable number of the coefficients are coprime with p. Choose iy so that plj || Cj and choose hj, lj so that Xj = hjk + \- and 0 ^ \] < k. Then F(x1? . . ., xs) = G(phixu ..., phsxs) where G(xx, . . . , xs) = d1pllx\ +... + dsplsxks with dj = CjPjXj. Now let h = max hy Then F(p*-*%,..., p*-fc-xs) = pfckG(x1,...,xs) and, for t > h, pt - h + hi pt - h + hs MF(P')> I ... Z 1 X! = 1 Xs = 1 phkG(x,, ..,xs) = 0(modp') >MC(p'"*k)n p j=i hk- h + h j whence Mpip^^Mcip1'^). (9.4) The form G can be rewritten as G = G(0) + pG(1) + . .. + ^-^-^ where fc Go-> = GU)(XU-)) = £ ^X*. I = 1 ', = J Clearly there exist i and r with r ^ s/k and G(0 containing at least r variables. Consider the form H(xl9 . . . , xs) = ( £ pJGiS)(px{S)) + X pJGiJ\xiJ)) V \ Now Mc(pr)^Mw(^-') (9.5) and H has the shape H = H{0) + pH(1) + . . . + pk~ lH(k- l) with H{0) containing at least r variables, where r ^ s/k, and all its coefficients relatively prime to p. It can be assumed, if necessary by
150 Homogeneous equations and Birch's theorem relabelling the variables, that H(0) = /j<°>(Xl,. . . , xr) = dxx\ + ... + drxk. By (9.5) and (9.4), to prove (9.3) it now suffices to show that there is a positive number C1{p) such that for t sufficiently large MH(pt)>Cl(p)p^~l\ (9.6) Let t denote the highest power of p dividing k and write y = z + 1 when p > 2 or t = 0, and y = t + 2 when p = 2 and t ^ 1. Then, as in §26, (9.6) will follow on showing that, for each m, d^x\ +...+ drxkr = m(mod py) (9.7) is soluble in x1,. . . , xr with p\xv Let K = py~x~ l(k, px(p — 1)). Then the number of /cth power residues modulo py is (p(py)/K. Hence, by Lemma 2.14, the set M ■ of residues m modulo py which can be written in the form dxx\ + ... +djXkj (pJlXi) satisfies card^> min{pyJ4k, i.e. s > 4/c2 —/c, then (9.7) has a solution of the desired kind, and this completes the proof of Theorem 9.1. Suppose that cl9. . . , cs are integers such that for every q the congruence c^ + . . . + csxks = 0(mod q) has a solution with (xj9 q) = 1 for some/ Then, following Davenport & Lewis (1963) c1?. . . , cs are said to satisfy the congruence condition. They define T*(/c) to be the least s such that every set of s integers c1?. . . , cs satisfies the congruence condition. They further define G*(/c) to be the least number t such that whenever s ^ t the equation cxx\ + . . . + csxks =0 has a non-trivial solution in integers when cx,..., cs are not all of the same sign when k is even and satisfy the congruence condition. The argument above gives T*(/c)^ 4/c2 — k + 1 and G*(/c)^ min(s0, 2k + 1). Davenport and Lewis show (i) that T*(/c)^/c2 + 1, (ii) that r*(fe) = k2 + 1 when k + 1 is prime, and (iii) that G*(fe) ^ k2 + 1 when/c > 18and/c ^ 6. Vaughan (1977b, 1989a) has removed the gap in (iii) by using the methods of Chapters 5, 6, 7 and 12.
Birch's theorem 151 For small values of/c, T*(/c) is known. (See Bierstedt (1963), Bovey (1974), Dodson (1967), Norton (1966).) Also, following earlier work of Norton (1966) and Chowla & Shimura (1963), Tietavainen (1971) has shown that r*(2fe+l) 2 lim sup —— = . k->oc /clog/c log 2 9.3 Birch's theorem Theorem 9.2 (Birch, 1957) Let j, I denote natural numbers and let /c1? . . . , kj be odd natural numbers. Then there exists a number T-(fcl9 ..., fc-, I) with the following property. Let Fx(x), ..., Fj(x) denote forms of degrees /c1? . . . , kj respectively in x = (xl9 . . . , xs) with rational coefficients. Then, whenever ^4^,...,/^,/) there is an l-dimensional vector space V in Qs such that for every xeV Fl(x) = ... = Fj(x) = 0. The first step in the proof is to establish the case when j = 1, F x is additive and k > 3. Lemma 9.1 There is a number <£(/c, I), defined for natural numbers k, I with k odd and k ^ 3, such that, if s ^ 0(/c, /), then for each form cxx\ + . . . + csxks with cx, . . ., cs rational, there is an l-dimensional vector space V in Qs such that for every xeV cxx\+... + csxks=0. (9.8) Proof By Theorem 9.1 there are t = t(k) and yx, . . ., yt not all zero such that cxy\ +. .. + ctyk = 0. Similarly for and so on. Hence, when s ^ It, the point (uxy1, . . . , u^yt, u2yt+ !,..., u2y2v • • • ■> ui);it-> v), . . . , 0) satisfies (9.8) for all ux,. . . , ut.
152 Homogeneous equations and Birch's theorem Proof of Theorem 9.2 Let A: = max/c,-, so that k is an odd positive integer. The proof is by induction through odd values of k. The result for k = \ is straightforward. For k ^ 3 the principal step is to show that if the result holds for systems of forms with max k( ^ k — 2, then it holds for a single form of degree k. The conclusion is then easily extended to a system of forms of degree at most k. For a form F(x) = F(x1,. . . ,.\\s)= ^ <'.-, .vri,> = I uj,---»jk i <■„ >jV ■ ■ ■ c■ ji ./k «1 «k 0 < ./,. < H + 1 Now define e{1) = (1, 0, 0, . . .), e(2) = (0, 1, 0, . . .), and so on, and take u0 = u, j(0) = j, y n = e(1), v(2) = e{1\ .... Then a further regrouping of terms gives F(vy + ule{l) + . . . + un + xe{n+ 1]) k = Z vh Z Uh--uh-kF(y'^h,--*h-h) (99) h = 0 j\....Jk-h 1 < j,. < fl + 1 where F(y;hJ1 Jk-h) is a form of degree h in j = (yx,. . . , vs). The total number of such forms with h odd, 1 ^ h < /c — 2 and 1 ^yr ^ n + 1 does not exceed k(n + l)fc. Hence, on the inductive hypothesis, and provided that s > %(n+1)fc(fc-2,...,/c-2,1), one finds that the corresponding simultaneous equations F(y*Aj1,...Jk-h) = 0 have a non-trivial solution z{0) in Qs. If z(0\ e( l \ . . . , e(" + l > are linearly dependent over Q, then omitting
Birch's theorem 153 one of the e{j) gives a linearly independent set of n + 1 points in Qs. Thus, in any case, by taking one of the uj to be zero in (9.9) and, if necessary, relabelling, one obtains z(0), z(l\ ..., z(n) that are linearly independent and such that k - 1 F(vz{0) + ulzil) + ... + unz(n))=cvk+ ^ vhGh(u) + G0(u) (9.10) /i = 2 h even where Gh(u) is a form of degree k — h in u = (wp . . . , w„). The linear independence of z{0K..., z{n) ensures that, when x = vzi0) + u1zn) + ... + unz{n\ non-trivial choices for (i\ ux,. . . , un) give non-trivial values for x. Consider the system of forms G„(m) = 0, /?even, 2^/?^/c-l. (9.11) The degree, k — h, is odd in each case. Hence, a further application of the inductive hypothesis shows that when n ^ 4/fc(/c — 2,. . ., k — 2, m), i.e. .s^.s0(/c, m), the system (9.11) is soluble for every member u of an m-dimensional vector space U in Q". Let w(1>, . . . , w(m> denote m linearly independent points in U and consider u = wxu{l) + . . . + wmu{m\ The linear independence again ensures that non-trivial w in Qw give rise to non-trivial u in Q". Hence, by (9.10), for non-trivial (r, u'j,. . . , wwi) there are non-trivial jc = (x„ . . . , xs) such that F(jc) = c^ + H(w) where H is a form in w = (w1}..., wm) of degree /c, i.e. F represents ci;fc + //(w). Continued repetition of this argument shows that if s^s{(k, /), then F represents a diagonal form cxv\ + . . . + ctvkt with t = 0(/c, /). Lemma 9.1 now gives the case j = 1, k1 = k of the theorem. To complete the inductive argument, it remains to establish the general case of j simultaneous equations Fx = . . . = Fj = 0 with
154 Homogeneous equations and Birch's theorem max fc, = fc. This is done by subinduction on j. The case 7 = 1 has just been dealt with. Suppose j > 1. Without loss of generality it can be supposed that fc ■ = k. By the case 7 = 1, given m, if s > *¥i(kp m), then there is an m-dimensional vector space U in Qs such that Fj(x) = 0 for every jc in U. The points of U can be represented by where Jt(1),. . . , x{m) are linearly independent points of Q\ For these points the forms F1?. . . , F}_ x become forms in y = (y1, . . . , ym). If max kt < k — 2 1 < 1 < j - 1 then one uses the main inductive hypothesis. If max fc,- = fc 1 < i < j - 1 then one uses instead the subinductive hypothesis. In either case, provided that m ^ 4^ _ 1(kl,. . . , fc;_ 1? /), there is an /-dimensional vector space V in Qm on which each Ft vanishes. This completes the proof of the theorem. 9.4 Exercises 1 Adapt the methods of Chapter 7 to show that limsup — -^ 2. fc^oc fclogfc 2 Adapt the methods of Chapter 6 to show that G*(3)^8, G*(4) ^ 14, G*(5) ^ 23, G*(6) < 36. 3 Show that r*(2) = 5, r*(3) = 7, r*(4)=17, and that r*(fc)^min(50,2fc + l).

10 A theorem of Roth 10.1 Introduction van der Waerden (1927) proved that given natural numbers /, r there exists an n0(l, r) such that if n > n0(l, r) and {1, 2,. . . , n} is partitioned into r sets, then at least one set contains / terms in arithmetic progression. For an arbitrary set ja/ of natural numbers, let A(n) = A(n,^/)= £ I D(n) = D(n, rf) = -A(n) (10.1) a < ii H ae, 0 contains arbitrarily long arithmetic progressions. An equivalent assertion is that if there is an / such that s4 contains no arithmetic progression of / terms, then d(*t) = 0. The first non-trivial case is 1 = 3. The initial breakthrough was made by Roth (1952, 1953, 1954) in establishing this case by an ingenious adaptation of the Hardy-Littlewood method. By a different method, Szemeredi (1969) proved the conjecture for / = 4, and Roth (1972) has given an alternative proof by an approach related to that of his earlier method. In 1975, Szemeredi established the general case. Unfortunately Szemeredi's proof uses van der Waerden's theorem. More recently Furstenberg (1977) has given a proof of Szemeredi's theorem based on ideas from ergodic theory. Although this does not use van der Waerden's theorem it apparently has a similar structure and so still does not yield the sought after insight.

156 A Theorem of Roth Ideas stemming from the attacks on this problem have enabled Furstenberg (1977) and Sarkozy (1978a, b) to establish that if d{srf) > 0, then the set of numbers of the forma —a' withaeja/, a'eja/ contains infinitely many perfect squares. In this chapter, Roth's theorem is established using his version of the Hardy-Littlewood method, and a proof of the Sarkozy- Furstenberg theorem is developed along the lines of Furstenberg but avoiding the ergodic theory. Throughout this chapter implicit constants are absolute. 10.2 Roth's theorem Let Mil)(n) denote the largest number of elements which can be taken from {1, 2,. . . , n} with no / of them in progression. Let Then Szemeredi's theorem is the assertion lim„^ ^ fi{l)(n) = 0, and this obviously implies the Erdos-Turan conjecture. As the following lemma shows, it is quite easy to prove that the limit exists. Its value is another matter. Lemma 10.1 For each integer /, lim„ _ a fi{l)(n) exists. Also, for m > n one has fi{l)(m) < 2fiil){n). Proof It is a trivial consequence of the definition of M(Z) that M{l)(m + n) ^ M(Z)(m) + M(l)(n). Hence Mil){m)^ m n M{l){n) + Mil)(m-n m n YYl ^-M{l)(n) + n n Therefore fi{l)(m) ^ fi{l)(n) + n/m, so that lim sup fiil)(m) < fi{l){n) m -* oc whence lim sup fiil){m) < lim inf fi{l)(n). m -* oo n -*■ oc Also, when m > n, M{l)(m) < (m/n + l)M{l)(n) ^ 2M{l)(n)m/n.
Roth's theorem 157 The following theorem not only shows that when / = 3 the limit is 0, but gives a bound for the size of M{3)(n). Theorem 10.1 (Roth) Letn>3. Then fi{3)(n) < (loglogn)" K It is henceforward supposed that I = 3, and for convenience the superscript (/) is dropped. Choose M cz {1, 2,. . . , n) so that card^ = M(n) and no three elements of Ji are in progression. Let m € ,Al Then a M(n) /(a)2/(-2a)da (10.3) o since the right-hand side is the number of solutions of m1 + m2 = 2m3 with m-eJi and, by the construction of Ji, such solutions can only occur when m1 —m2 — ra3. Let k denote the characteristic function of Ji, so that /(a) = 5>(xMa*). (10.4) Suppose that and consider m < n, (10.5) v(oL) — fi(m) ]T e(ax) (10.6) X = 1 and Then E(a) = i;(a)-/(a). £(a) = X c(*)e(ax) (10.7) X = 1 with c(x) = fi(m) - k(x). (10.8) The idea of the proof is that, if M(n) is close to n, then 2 /(a)2/(-2a)da 0
158 A Theorem of Roth ought to be closer to M(n)2 than to M(n)(d. (10.3)). To show this, one first of all uses the disorderly arithmetical structure of Ji to replace f by v with a relatively small error. It is a fairly general principle, observable from the applications of the method in previous chapters, that sums of the form Z e(*x) x ^ n X€£j/ tend to have large peaks at a/q when the elements of srf are regularly distributed in residue classes modulo q. Note that v(ol) has its peaks at the integers. Let m - 1 F(a)= X ^(az). (10.9) = = o Lemma 10.2 Let q be a natural number with q < n/m, and for y = 1, 2, . . . , n — mq let m - 1 Q (y =1,2,...,n-mq) (10.11) and n — mq F((xq)E((x)= £ (T(y)e(a(y + mq-q)) + R((x) (10.12) y= i where R(cc) satisfies \R((x)\<2m2q. (10.13) Proof By collecting together the terms in the product FE for which x + zq = /7 + mq — q one obtains n F(aq)E(cc)= Yj e{Roth's theorem 159 contribution from the terms with h ^ 0 and h> n — mq does not exceed, in modulus, m(mq+(m—l)q) <2m2q. For the remaining values of h one has 1 < h + q(m — 1 — z) ^ n for all z in the interval [0, m - 1]. This gives (10.12) and (10.13). By (10.8) and (10.10), m - 1 + * £(0)- £ ifi{m) - k(x)) = n(n(m) - n(n)). x= 1
160 A Theorem of Roth The lemma follows at once. Proof of Theorem 10.1. Let '•i I Then, by (10.4) and (10.6), /(a)2r(-2a)da. (10.14) o / = I I /*(»»)• ae M be.At 2\a + b Thus, if Mj is the number of odd elements of M and M2 the number of even elements, so that M1 + M2 = M(n\ then / = n{m){M\ + M\) > ^(m)M(n)2. (10.15) By (10.3) and (10.14), |/(a)|2da. |M(n)-/|^(max|£(a)| o Therefore, by Lemma 10.3 and Parseval's identity, when 2m2 < n one has \M(n) -/I** (2n(fji(m) - ft(n)) + 16m2)M(n). Hence, by (10.15), fi(m)fi(n) <: 4(fi(m) - fi(n)) + 34m2n~ x (2m2 < n). (10.16) Letting n—> oo and then m—► oo shows that t = lim„_+ ^/^(^) satisfies t2 ^ 0. To establish the quantitative version of this, let X(x) = fi(23Xy By Lemma 10.1, it suffices to show that A(2x) <^x~ 1. By (10.16), A(y)A(y + 1) ^ 4(A(y) - A(y + 1)) + 34 x 2" 3>. Dividing by A(_y)A(_y + 1), summing over j/ = x, x + 1,. . . , 2x — 1 and appealing to Lemma 10.1 gives one x < 4A(2x)" x + 200xA(2x)" 22~ 3\ When /l(2x) > 1/x the second term on the right is < \x for x sufficiently large, so that /l(2x) < 8/x, which gives the desired conclusion.
A theorem of Furstenberg and Sarkozy 10.3 A theorem of Furstenberg and Sarkozy 161 Theorem 10.2 Let srf bea set of natural numbers with d(srf) > 0, and let R(n) denote the number of solutions of a—a' = x2 in a, a', x with aesrf, a'estf, a^n. Then \imsupR(n)n~3/2>0. n -»• oo This theorem is somewhat stronger than Theorem 1.2 of Furstenberg (1977). The approach of Sarkozy (1978) is different. He adapts the methods of § 10.2 to show that ifa — a' = x2 has only trivial solutions, then A(n) < n(\og log n)2/3(logn)~ 1/3. Let *y0 denote an infinite set of natural numbers such that lim n~ M(n) = 3(j?/), n -*■ oo let and let Wln(q, a) = {a : |a - a/q\ ^ q~ ln~ 1/2], (10.17) /(a)- £ e(a.a). a < n aesf It is necessary to show that f has fairly orderly behaviour on Wln(q, a). For n ^ 4, /(a)|2da^ Wln(q,a) |/(a)|2da ^ n. 0 Hence \f(a)\2n~lda Wln(q,a) is bounded uniformly in q, a, n. Therefore one may choose infinite sets J^(q, a) of natural numbers such that jV(1, 1) = ^"(1, 0) c.V09 ¥{q + 1, 1) c f(q, q - 1),
162 A Theorem of Roth Jf(q, a') cz ^V(q, a) when l^a q = Q + 1 a = 1 («,«) = 1 Now define /c = (g!)2, P = k100 and henceforth suppose that neJ^(P, P- 1). Then, given choose n0 = n0(>/,X)^X2 1 (10.18) (10.19)
The definition of major and minor arcs 163 so that when n^n o and 1 ^ a ^ q ^ P with (a, q) = 1 the major arcs SRn.xfo fl) = {« : |a - a/q\ < Xq~ ln~ l} are pairwise disjoint, |/(a)|2tt Ma < p(q, a) + f]P (10.20) 9Wr,(4,fl) and 4(n)>fdn. (10.21) By (10.17), 9K„>X(^, a) c <3Rn(q, a) for n 3* n0. Hence, by (10.20), |/(a)|2n_1da
a) Let 5)¾ denote the union of the major arcs WlntX{q,a) with 1 ^ a ^ q < P and (a, g) = 1, and define the minor arcs m by m = {Xn~\l + Xn-^\m. Further define N g(fl= I e(Px2) x= 1 (10.23) where N^(n/k)1/2. (10.24) Then, by (10.19), R{n)^0t where By Theorem 4.1, when (a, g) = 1 n <,(/ca)|/(a)|2da. (10.25) o a g(y) = q~lS(q9 a)h[y--) + 0[ q9,16( 1 + N2 y a q (10.26) where S(q,a)= £ e(ax2/164 A Theorem of Roth 10.5 The contribution from the minor arcs Suppose aem. Choose a, q so that (a, g)=l, g P or |a —a1/q1\> Xn~ 1q^ \ In the first case, by (10.26), g(koc) Xn~ 1q~ 1, so that, by (10.26), (10.27) and Lemma 2.8, 1/2 g(ka). (10.28) m 10.6 The contribution from the major arcs Now suppose that ae^)ln x(q, a) where 1 ^a < q < P and (a, q) = 1 Let qx = q/(q, k\ ax =ak/(q, k). Then, by (10.26), g(ka) = q1 ^(q^ajhlklct a q. a a q + Olq91/16U+N2k The error term here is majorized by P + N2kXn~ *. Hence g(ka)\f((x)\2d(x = MX+ 0(Pn + N2kX) (10.29) 2R where *i= I I q^P a= 1 (a,4) = 1 a ^^(^fli)/! fc a-=))|/(a)|2da.
Completion of the proof of Theorem 10.2 165 By (10.22) and (10.18) the terms here with q ^ Q + 1 contribute an amount which in absolute value does not exceed P q £ X Nn(p(q,a) + r]P~2)<2r]Nn. q = Q + 1 a= 1 ia,q)= 1 Also, when q ^ Q, by (10.19) one has g|/c, so that q1 = 1. Hence >2 + OfoAfa) (10.30) where I I hi ki a q^Q a= 1 Jan„,x(q,a) (a, 4)= 1 a q. ||/(a)|2da. It is easily shown that for every positive number Y a~ 1/2cosada > 0. o Hence, by (10.27), Re/iOS)HjT1/2 rN2\p\ k(x. 1/2cos27rada > 0. (10.31) 0 Therefore, on discarding all the terms in 0t2 with the exception of that with a = q = 1, one obtains ri/4nn Re@i> Re/i(/ea)|/(a)|2da. 1/47TH Also, when |a| ^ l/(4nn), one has /(a) -/(0)= £ 27cix so that x= 1 X6 J2/ ■a e(Px)dp o |/(a)| > /(0)(1 - 27r|a|n) > ±/(0) = ^(n). Therefore ReM2^±A(n): (•1/4-nn Re/i(/ea)da. (10.32) o 10.7 Completion of the proof of Theorem 10.2 By (10.24) and (10.31), Re fc(fca) ^ \N whenever 4nn\(x\ ^ 1. Hence, by (10.32) and (10.21), 32nn 250
166 A Theorem of Roth Thus, by (10.25), (10.28) and (10.30), R(n)>@ = Re@ = Re^2 + 0((Nk~40 + n1/2X~ 1/2 + N3/4)n +rjNn) d2 > nN - C((Nk~ 40 + n1/2X~ 1/2 + N3/*)n + rjNn) 250 (10.33) for a suitable constant C ^ 1. The proof is completed by making suitable choices of the parameters. Let *y= 10" 4a2C_1. This fixes Q = Q0(rj) and so k and P. Note that, by (10.18) and (10.19), k>Q>l/ti. Let X = rj-2k and suppose that n^n0(rj, X) with neJr(P, P— 1). Finally, let N = [(n/fc)1/2], so that (10.24) holds. Now for n ^ n^), C((Nk-*° + n1/2X" 1/2 + N3/4)n + yyNn) < C(rjNn + rjn3/2k~ 1/2 + f/Nn + ^iVn) < 5CrjNn d2 Nn. 2000 Hence, by (10.33), lim sup R(n)n~ 3/2 ^ 32/c" 1/2 > 0 n -* oo 3UU as required. 10.8 Exercises 1 Prove the theorem of Sarkozy stated in § 10.3. 2 Show that if d(s/) > 0 and R(n) denotes the number of solutions of a —a' = p — 1 with aestf, a'estf, a^n, p prime, then limsupJR(n)(logn)n 2 > 0. n -* oo
11 Diophantine inequalities 11.1 A theorem of Davenport and Heilbronn All of the forms of the Hardy-Littlewood method described so far have dealt with the solution of equations in integers. For instance, in Chapter 9 it was shown that if s is large enough, then given integers cl9 . . ., cs (or equivalently given that c1?. . ., cs are all in rational ratio), not all of the same sign when k is even, the equation cxx\+.. .+ csx* = 0 has a non-trivial solution in integers x1,.. ., xs. Now one can ask what happens when the c1?. . ., cs are not in rational ratio. It is no longer sensible to insist that the form represents 0, but one can ask instead that it take arbitrarily small values. In order to answer this question, Davenport & Heilbronn (1946) introduced an important variant of the Hardy-Littlewood method. This enabled them to establish the following theorem. Theorem 11.1 Suppose that s ^ 2k + 1 and tlmt /^,..., Xsare non-zero real numbers not all in rational ratio, and not all of the same sign wlien k is even. Then for every positive number n there exist integers x1? . . . , xs, not all zero, such that 1^+...+^1^. (11.1) It suffices to prove the theorem when n = 1, for it can then be applied with Aj replaced by Aj/rj. Moreover, when k is odd, replacing, if necessary, x\ by ( — x^f enables one to assume in this case also that not all the Aj are of the same sign. By relabelling it can be supposed that kJX2 is irrational. If kJX2 > 0, then consider any j for which ^Jkj < 0. Then, when a^a- is rational, A2/A7 is irrational and negative. In any case, by further relabelling it can be supposed that a1/a2 is irrational and negative. (11.2) In all of the forms of the Hardy-Littlewood method considered so
168 Diophantine inequalities far, the line of attack has been via fourier transforms on the torus T = M/Z. For the present problem it is more appropriate to work on U. The obvious analogue of (1.8) is the identity. <•* / ^sin27ca fl (|jB| < 1), eioLp)— da = < . - x ^ 10 (|/J| > I)- However there are difficulties associated with this transform because the integral does not converge absolutely. It is more convenient, therefore, to use instead /(/0 = 1 *- /sin7iax e(aj8)K(a)da, K(ol)=[ . (11.3) , ticl oc \ A straightforward application of the Cauchy integral formula gives /(/*) = max(l-101,0). (11.4) Lei /(«)= £ e(axk), fj(oi) = f{Aja). (11.5) X = 1 Then for the method to be successful one requires a positive lower bound for R(N) = oc 11 fj(a)\K(a)da, (11.6) - oc \j = 1 / for by (11.4) and (11.5) this is N N £ ••• L max(1-1^+...+^1,0) Xi = 1 xs = 1 which can only be positive if there are x ^,. . . , xs for which (11.1) holds with r\ = 1. Thus Theorem 11.1 follows from Theorem 11.2 Suppose that s > 2k. Then there are arbitrarily large N for which R(N)>Ns~k. Note that throughout this chapter implicit constants may depend on /1? . . ., /s. 11.2 The definition of major and minor arcs The form of the Hardy-Littlewood method used here is somewhat simpler than that described hitherto. The most important
The treatment of the minor arcs 169 simplification arises from the fact that for suitable choices of N the integrand has only one really big peak, that at the origin. This is because the irrationality ofk1/A2 ensures that one of/^ ,/2 is relatively small when a is not near the origin. Let v = 4 P = N«. (11.7) Then R is divided into three regions. These consist of the sole major arc m = {a:\a\^PN-k}, (11.8) the pair of minor arcs m = {a:PAr*<|a|
P}. (11.10) The trivial region can be dismissed quickly. By Hua's lemma (Lemma 2.5), rx +1 \2k \fj(*)rd*170 Diophantine inequalities Lemma 11.1 Let a, q be any pair with (a, q) = 1 and /lx a k2 q 0. N-+ oo N m Proof of Lemma 11.1 Suppose that N ^ N0(/l9... ,/s), let aem and Q = Nk~v/2 and choose qj9 ap in accordance with Lemma 2.1, so that (qj9 aj) = 1, qj < g, \kp - aj/qjl < l/(qjQl The first step is to show that at least one of ql9 q2 is relatively large. If a-} were to be 0, then one would have \oc\^l/(qjQ\kj\)The treatment of the minor arcs 171 On hypothesis, l2 q with |0|<1. Hence a q2a1 - 1 _ „- 2 <^e~ + anc* if it is zero, then a/q = (q2a1)/(a2q1), which again implies that \a2qi\f>q- Since a2 = X2VLq2 — 02Q~ 1 <^ q2P it follows that q1q2 f> qP~ *. Therefore, by (11.7), max(^1?^2)>N1/5. (11.13) Now, by Weyl's inequality (Lemma 2.4), for j = 1, 2, fj{*) < N 1 +£ 1 1 , *1 21 ~fc + TT + Hence, by (11.13), minfl/^UAWIMW1-* as required For the remainder of the proof of Theorem 11.2 it will be assumed that N is chosen in accordance with the specialization given in Lemma 11.1. Let m1 = {a :aem, 1/^ (a)| < |/2(a)|}, m2 = m \m1. By (11.3), X(a) <^min(l, a"2). Also the argument that gives (11.11) can be readily adapted to show that rx + i 2*+ 1 n /A*) i = 1 da<^N 2k-k +£ so that m 2k + 1 i = i X(a)da ^ JV 2k-k + £
172 Diophantine inequalities Therefore, by Lemma 11.1, when; = 1 or 2, 2k + 1 n /•(«) Jm i = 1 X(a)da < N 2k+l-k-d + e Thus, there is a positive number 3 such that X(a)da^Ns-fc-^. Jm n m j= i (11.14) 11.4 The major arc In view of (11.12) and (11.14) it remains only to show that, for N sufficiently large, [I fj(*))K(*)daL>N an \j = l s- k (11.15) Let a6¾)¾. By (11.7), (11.8), Lemma 2.7, and the remark after the proof of that lemma, one has /)(01) = Vj(a) + 0(N2V) where e(XjThe major arc 173 Hence, by (11.7) and (11.8), f] v(ol) )K((x)d(x ^ 00 a"s/fcda Nv-k ^ ^y(s-fc)(l-v/fc) Thus 17 ^(a))X(a)da = OR V/ = 1 '00 n Vj(ol) )K(a)da - 00 \j = 1 + 0(Ns-fc-*). (11.18) By (11.16), '00 n Vj{(x) )K(a)da - 00 \j = 1 ■00 rN da — 00 ,/ /*JV dp!.. 0 e((X&+... + Aj?»K(a)dft 0 Since K(a) <^ min (1, a 2) and the integrand is continuous the order of integration may be interchanged. Hence, by (11.3) and (11.4), '00 f] Vj{ol) )K(a)da -00 \j = 1 rN rN dp, o max(l-^/¾+... +A,^|,0)d/S, 0 rN* = k — s rN* dcc1 o (a,... a,)1""1 0 x max(l — \^1oc1 +... + /lsas|, 0)das. It is now that one requires the hypothesis that ^1/^2 < 0- Consider the region & = {(a2,. . . as): 3Nk ^ a2 ^ 23N\ 32Nk ^ a,. ^ 232Nk (3 ^ ; ^ s)}. Then, for 3 sufficiently small, whenever (a2,..., cts)e& one has 232Nk < - {k20L2 + ...+ Asas)A" l < \Nk and so every cc1 with lA^ +...+ AsaJ ^ \ satisfies 32Nk (Nl-kY -00 V/ = 1 da-. . . . da. dax j2/(a2,...,as)
174 Diophantine inequalities where s#(ct2,. .. , as) denotes the interval with end points (— (A2 a2 + ... + Xsas) + |)/l [" *. Obviously the volume of 0& is ^iV^-1. Hence Too / s \ - ao\j= 1 J This with (11.17) and (11.18) establishes (11.15), and thus completes the proof of Theorem 11.2. 11.5 Exercises 1 (Davenport & Roth, 1955; Vaughan, 1974b) Obtain Theorem 11.1 for any s^C/clog/c where C is a suitable constant. 2 Let /l1? /l2, /l3, /i, n denote real numbers with X} =/= 0, rj > 0, Xl/X2 irrational, and kjk2 < 0. Show that there are primes pl9 p2, p3 such that 3 (Baker, 1967; Vaughan, 1914a) Modify the argument used to answer 2 above so as to show that there are infinitely many triples of primes pl9 p2, p3 such that l^iPi + ^iPi + ^3^3 + lA < (log maxpj)~ \ j 4 (Baker)."^" Let F(N)-+0 as N-+oo. Prove that the statement 'for every sufficiently large N there are primes p1, p2, p3 such that pj ^ N and I X1p1 + X2p2 + /l3Jp31 < F(N)' can be false for suitable Al9 A2, A3 with Ai/^2 > 0 and A.JA.2 irrational. tCommunicated in conversation in June 1973.