KLOOSTERMAN SUMS WITH PRIMES AND THE SOLVABILITY OF ONE CONGRUENCE WITH INVERSE RESIDUES - II Текст научной статьи по специальности «Математика»

ЧЕБЫШЕВСКИЙ СБОРНИК

Том 21. Выпуск 1.

УДК 511.321 DOI 10.22405/2226-8383-2020-21-1-221-232

Суммы Клоостермана по простым числам и разрешимость одного сравнения с обратными вычетами — II

М. А. Королев

Королев Максим Александрович — доктор физико-математических наук, ведущий научный сотрудник, Математический институт им. В. А. Стеклова РАН (г. Москва). e-mail: korolevma@mi-ras.ru

Аннотация

В настоящей статье продолжены исследования, связанные с распределением обратных вычетов по заданному модулю. Ранее автором был получен ряд нетривиальных оценок коротких сумм Клоостермана с простыми числами, отвечающих произвольному модулю q. Следствием таких оценок стали результаты о распределении вычетов р, обратных простым числам "короткого" промежутка: рр = 1 (mod q), 1 < р < N, N < q1-S, 5 > 0, и, более общо, о распределении по модулю q величин д(р) = ар+Ьр, где а, Ь - целые числа, (ab, q) = 1.

Еще одно приложение найденных оценок связано с задачей о представимости произвольного заданного вычета m (mod q) суммою д(р\) + ... + д(рк) при фиксированных а, Ь и к ^ 3, и простых 1 < pi,... ,рк ^ N. Для количества таких представлений автором была найдена формула, поведение предполагаемого главного члена которой определяется аналогом "сингулярного ряда" классического кругового метода, т.е. некоторой величиной к, зависящей от q и набора к, а, Ь, т. При фиксированных к, а, Ь, т она является мультипликативной функцией q. В случае, когда модуль q не делится на 2 или 3, эта величина строго положительна, так что формула для искомого числа представлений является асимптотической.

В настоящей работе исследуется поведение к в случае, когда q = 3". Оказывается, что при любых п ^ 1 k ^ 3 существуют "исключительные" тройки а,Ь,т, для которых к = 0. Цель работы состоит в описании всех таких троек и нижней оценки величины к для "неисключительных" троек.

Ключевые слова: сравнения, разрешимость, обратные вычеты, суммы Клоостермана, простые числа, сингулярный ряд.

Библиография: 17 названий. Для цитирования:

М. А. Королев. Суммы Клоостермана по простым числам и разрешимость одного сравнения с обратными вычетами — II // Чебышевский сборник, 2020, т. 21, вып. 1, с. 221-232.

CHEBYSHEVSKII SBORNIK Vol. 21. No. 1.

UDC 511.321 DOI 10.22405/2226-8383-2020-21-1-221-232

Kloosterman sums with primes and the solvability of one congruence with inverse residiiGS XX

M. A. Korolev

Koro lev Maxim Aleksandrovich — doctor of physical and mathematical sciences, the leading researcher of the Department of Number Theory of Steklov Mathematical Institute of RAS (Moscow).

e-mail: korolevma@mi-ras.ru

Abstract

In the paper, we continue to study the distribution of inverse residues to given modulus. Earlier, the author obtained a series of non-trivial estimates for incomplete Kloosterman sums over prime numbers with an arbitrary modulus q. One of the applications of such estimates are some assertions concerning the distribution of inverse residues p to prime numbers lying in a "short" segment: pp = 1 (mod q), 1 < p ^ N, N ^ q1-S, S > 0, and, more general, concerning the distribution of the quantities g(p) = dp + bp with respect to modulus q, where a, b are some integers, (ab, q) = 1.

Another application is connected with the problem of the representation of a given residue m (mod q) % the sum g(p1) +... + g(pk) for fixed a, ^d k > 3, in primes 1 < p1,... ,pu < N. For the number of such representations, the author have found the formula, where the behavior of the expected main term is controlled by some analogous of the "singular series" that appears in classical circle method, that is, by some function «depending on q and the tuple k, a, b, m. For fixed k, a, b, m, this function is multiplicative with respect to q. In the case when q is not divisible by 2 or 3, this function is strictly positive, and therefore the formula for the number of the representations becomes asymptotic.

In this paper, we study the behavior of « for q = 3". It appears that, for any n > 1, k > 3, there exist the "exceptional" triples a, b, m such that « = 0. The main purpose is to describe all such triples and to obtain the lower estimate for « for all non-exceptional triples.

Keywords: congruences, solvability, inverse residues, Kloosterman sums, prime numbers, singular series.

Bibliography: 17 titles. For citation:

M. A. Korolev, 2020, "Kloosterman sums with primes and the solvability of one congruence with inverse residues — II Chebyshevskii sbornik, vol. 21, no. 1, pp. 221-232.

To academician V. P. Platonov in occasion with his 80th anniversary.

1This work was supported by the Russian Science Foundation under grant 19-11-00001.

1. Introduction

In the present paper, we continue the study of the solvability of some congruences with inverse residues to modulo q started in fl] and [2]. As in [2], the main subject now is the congruence

g(pi) + ... + g(pk) = m (mod q), (1)

where q ^ 2 is an arbitrary integer, g(x) = ax + bx (mod q), k,a,b,m are any fixed integers satisfying the conditions k ^ 3, 1 ^ a,b,m ^ q., (ab, q) = 1. The variables p1,... ,pk run through prime numbers from the interval (1,^j. This interval is assumed to be "short", that is, we are interested in the case when N ^ q1-S to some positive 5.

The key role in such problems is played by the estimates of Kloosterman sums with primes, that is, of the exponential sums of the type

Wq(a,b; X) = ^ exp ^^^(ap + bp)j ,

p^X, p\q

(for such estimates and their applications, see: [3]-[16]). The estimates given in fl], [2] lead to the following assertion (see [2]):

Th eorem A. Let 0 < £ < 0.01 be an arbitrary fixed constant and let k ^ 3 be any fixed integer. Suppose that q ^ q0(e,k). Further, let (ab,q) = 1 and g(x) = ax + bx (mod q). Finally, let

2(k + 33) ,t 3k + 50

Ik = —;-r- if 3 ^ k ^ 16 and 7k = ~m-r v % ^ 17,

№ 3k + 64 J m 4(k + 12) J '

and suppose that qlk +£ ^ N ^ g. Then the number Ik(N) = Ik(N,q,a,b,m) of solutions of (1) in primes pj ^ N, (pj ,q) = 1, satisfies the relation

Ik(N) = ^p (xk(q) + 0(Ak)). (2)

Here Kk(q) = Kk(a,b,m; q) is some non-negative multiplicative function of q for any fixed tuple k, a, b and m. Moreover,

a) for any k ^ 7 we have Ak = (lnlnN)B(lnN)-A,

1 29

A =2 + T(fc - 7), B = 2 - 1;

b) for any k ^ 3 we have Ak = q -£, if Generalized Riemann hypothesis is true. Remark. One can check that 7k ^ 1 — 733 to anv k ^ 3.

The ascertaining of the conditions of when (2) becomes the asymptotic formula is connected with the detailed study of the multiplicative function Kk(q) = Xk(a, b, m; q). In this direction, in [2], we prove the following assertion:

Theorem B. Suppose that q is coprime to 6) and Iet k ^ 3 be any fixed integer. Then, for any triple (a, b, m) with the conditions 1 ^ a,b,m ^ q, (ab, q) = 1 the following inequalities hold:

кк (a,b,m; q) ^

clexpJ, = 3,

сз(1п1пq)-6, if к = 4,

10 5, for any к ^ 5

where the constants Cj, j = 1, 2, 3; are absolute.

Remark. The exponent (—6) (for k = 4) and constant 10-5 (for k ^ 5) are not optimal.

Now the purpose is to study the behavior of Kk (q) in the case when q = 3n. Unlike the case (q, 6) = 1, for any pair k, n there exists the set Qk (3n) of "exceptional triples" (a,b,m) such that Kk(a, b, m; 3n) = 0. In what follows, we shall refer these sets as "exceptional sets".

The main results of this paper is the description of all exceptional sets Qk(3n), k ^ 3, n ^ 1. Namely, we prove here the following assertion:

Theorem. In the cases k ^ 8, n ^ 1 and 3 ^ k ^ 7, n = 1, the set Qk(3n) consists of the triples satisfying the conditions

1 <a,b,m < 3n, (ab, 3) = 1, a + b = 0 (mod 3), m^ 0 (mod 3). (3)

In particular, Q(3 n)| =4 ■ 33(n-1).

In the case 3 ^ k ^ 7, the set, Qk(32) consists of the triples listed in (13)-(16); in particular, |Qk(32)| = 18(14 — k)-,

Finally, in the case 3 ^ k ^ 7, n ^ 3, the set Qk(3n) consists of the triples coinciding modulo 32 with the triples from the set Qk(32); in particular, |Qk(3n)| = 2 ■ 33n-4(14 — k).

At the same time, for any k ^ 3,n ^ 1 and for any "non-exceptional" triple (a,b,m) ^ Qk(3n) one has

Kk(a,b,m;3n) > —.

50

2. Complete Kloosterman sums to prime power moduli

In [2], we establish the following formula for Kk(pn) = Kk( a, b, m;pn): Kk (pn) = 1 + Ak (p) + Ak (p2) + ... + Ak (pn).

Here

1 -2^

Ak (pn) = Ak (a,b,m; pn) = e Pn Sk (fa,fb; pn),

and S(a, b; q) is a complete Kloosterman sum, that is,

S(a, b;q) = ^ exp (ax + bx)j .

(.x,q) = 1

To study the properties of Kk(q), we need some explicit expressions and the estimates for the quantities S (a, b; pn), Ak (a,b,m;3n), n ^ 2.

Lemma 2.1. Let, p ^ 3 and (ab,p) = 1. If ab is a quadratic non-residue modulo p then S(a, b; pn) = 0 for any n ^ 2. Otherwise, setting q for pn and v for any solution of the congruence ab = v2 (mod q), we have

{2Jq cos for even n,

2V^(f) co's (4f- + ^) for odd n,

where s = 0 for p = 1 (mod 4) and s = 1 for p = 3 (mod 4). For the proof, see [17].

Corollary. Under the conditions of Lemma 2.1, for any n ^ 2 and q = pn we have

\S(a, b;q)\ < 2^q. Lemma 2.2. For any a,b coprime to 3 the following equalities hold:

2, if a + b = 0 (mod 3),

i-

S(a, b;3) = i

-1, otherwiset

„ Î3(-1)6 cos £ (4a + b), ifb-a = 0 (mod 3), S (a, b; 32) = I 9

I 0, otherwise.

Proof. Setting w = e2ni/3 we get

'2, if a + b = 0 (mod 3),

S (a, b; 3) = ^Twax+bx = wa+b + w2(a+b) = J2,

x=1 I-1,

Further, setting x = y + 3z in the sum S (a, b; 32) we obtain

3 3

S (a, b; 32) = ¿' e2^ ^+by) £).

^ - E ^

y=i ¿=1

So, 5(a, b; 32) = 0 for b — а ф 0 (mod 3). Finally, if a = 6 + 3n for some integer n then

S(a, b; 32) = 3(e™+ et(5a+2&)) = 3e™ ^) cos *(4a + 9).

9

Since

2 ^3a + y) = 2 (3 b + y + = b + 2n = b (mod 2), we arrive at the assertion of the lemma. □

In this section, we use the explicit formulas for Kloosterman sums S(a, b; pn), p^ 3, n ^ 1 to provide the explicit formulas for the values

1 -2*ii™ Ak(pn) = Afc(a,b,m;pn) = ^y E e pU Sk(fa, fb;pn).

Lemma 2.3. Suppose that (ab, 3) = 1. If a + b = 0 (mod 3) then

, , , |2, when m ф 0 (mod 3), , . „. Ak(3) = ' ф o ( ) Ak(3ra) = 0 for n > 2.

I— 1, when тф 0 (mod 3);

If a + b ф 0 (mod 3) then

. /o4 (—1)k+m жт (—1)k 2жт

Ak(3) = --,—-,— cos- = —-,—T- cos-,

2k-1 3 2k-1 3

32

ak (32) = E'(—1)k6/ cos2^ cosk ^f(4a +

k

cos-- cos

9 9

proof. In the case a + b = 0 (mod 3), Lemma 2.2 implies that

a (3) = -fc ■2fc e^ i=i

-mf

2, when m = 0 (mod 3), —1, when m ^ 0 (mod 3).

ab

It is easy to check that the conditions a + b = 0 (mod 3) and J = —1 are equivalent, so we have Ak(3n) = 0 fa any n ^ 2 by Lemma 2.1. In the case a + b ^ 0 (mod 3), Lemma 2.2 implies:

a (3) = 2i Ew-m/ f=i

(—i)fc =

(-1) 2k

(cu-m + w-2m) =

(-1)"

k -Kim

~nm(„ 3

■Kim

3

2k

e """(e 3 + e 3 ) =

(-1)fc+m nm

--,-cos-.

2k 3

Finally we have

Afc (32)

Qk

, n mf ^f

6k (-i)kbf cos H. (4a + b) =

f=i

(-1)kbf cos

W cos2^ cos fc ^ (4a + b).

i =i

9

9

Lemma is proved. □

Lemma 2.4. Suppose that s ^ n ^ 2, k ^ 5. Then the following inequality holds:

|E-^ (3r )

r=n

<

2 • 3^+™-kn/2-1

1- 31-fc/2 .

(f)

Lemma 2.5. Suppose that n ^ 5. Then, for any s ^ n and for any a,b satisfying the condition

— ) = 1 the following inequality holds:

EA3(3r )

— ■ 3(8-ra)/2, when n = 0 (mod 2), < < 156

— ■ 3(7-ra)/2, when n = 1 (mod 2); 16

Lemma 2.6. Suppose that n ^ 4. Then

Ea4(3r )

< - 3

5—n

Lemma 2.7. Suppose n ^ 3. Then, for a ny s ^ n, the following inequalities hold:

35

■ 33(4-n)/2, when n = 0 (mod 2),

< I 416

< ^ 55

■ 3(11-3n)/2, when n = 1 (mod 2);

416

Ea5(3r)

r=n

Lemmas 2.4, 2.5, 2.6 and 2.7 are particular cases of Lemmas 6.1, 6.3-6.5 from [2], respectively.

1

3. The singular series (a,b,m; q) for q = 3n

In the cases q = 2n, 3n, n = 1,2,3,..., the behavior of the series Kk (a,b,m;q) is more sophisticated than in the case (q, 6) = 1. In particular, for "small" k and any n there exist some "exceptional" set Q k( (?) of triples (a,b,m) such that Kk(a,b,m) = 0. These exceptional sets can

= 3 n = 2 n

consider only the case p = 3 which is more easy.

Lemma 3.1. Suppose that ( ab, 3) = 1, a + b = 0 (mod 3). Then, for any n ^ 1 and q = 3n we have

, . . | 0, when m ^ 0 (mod 3), Kk(a, b, m;q) = <

I 3, when m = 0 (mod 3).

PROOF. Obviously, the conditions (ab, 3) = 1, a + b = 0 (mod 3) are equivalent to the condition

= —1- In view °f Lemma 2.1> in this case we have Kk(q) = 1 + Ak(3). By Lemma 2.3,

Ak (3) = — 1 for m ^ 0 (mod 3) and Ak(3) = 2 for m = 0 (mod 3). Thus lemma follows. □

Corollary. For any n ^ 1, the set, Qk(3n) contains all the triples (a,b,m) satisfying the following conditions: 1 ^ a,b,m ^ 3n; (ab, 3) = 1, a + b = 0 (mod 3) m ^ 0 (mod 3). The number of such triples is equal 4 ■ 33(n-1).

Lemma 3.2. Suppose that k ^ 8, (ab, 3) = 1 and a + b ^ 0 (mod 3). Then, for any n ^ 1;

q = 3n and for any m, 1 ^ m ^ q, one has Kk (a, b,m;q) > —.

50

proof. Suppose first that n = 1. Sine e x = x (mod 3) to any x £ Z* then g(x) becomes a linear function: g(x) = ax + bx = (a + b)x. Hence, the congruence (1) is equivalent to

( a + b)(x1 + ... + xk) = m (mod 3)

or to

x1 + ... + xk = ^ (mod 3), ^ = m(a + b)* = m(a + b) (mod 3). (4)

At the same time, the number of solutions (x1,..., xk) of (4) such that (xj, 3) = 1, j = 1,... ,k, is equal to

^(3) = !£ £ e2m3= iZf)V2mf =

C=1 X1,...,X]Z = 1 C=1 X=1

1 3 i 2 i

= 1 + u2c)k = 1(2k + (—1)k £w-CM) = 1 (2k + (—1)k5),

3 3 3

=1 =1

where ш = е2жг/3,

Thus we obtain

1-,

. 2, when m = 0 (mod 3),

=

—1, when m ф 0 (mod 3).

юк(3) / (—1)к6\ , , (—1)к$ 1 1 15

Vk(3) = -t—^- 1 + -—, кк (3) = 1 + -—^ 1 j г- ^ 1--= —.

3 ^ i 2к J ку ' 2к 2к-1 16 16

Further, let n = 2. Since j = 1, then Lemma 2.2 implies

32

Кк(32) = 1 + cos ^ + ¿'(—1)^ cos (^) сок (4a + 6). (5)

^ - 1 ' (—1) cos ^m + £'(—1)^ cos ^ —к

f=1

The direct tabulation over all triples ( a, b,m), 1 ^ a,b,m ^ 9, (ab, 3) = 1 shows that

Kk(32) > 0.0351562 ...for k = 8, 9, (6)

Kk(32) > 0.0966797 ...for k = 10,11, (7)

Kk(32) > 0.171387.. .for k = 12. (8)

Moreover, (4) implies the inequality

32

i o i 1 I 1 f k

I Kk(32) — 1 | < 2k + EI cos If(4° + b) .

The condition a + b ^ 0 (mod 3) implies that 4a + b ^ 0 (mod 3), that is, (4a + b, 3) = 1. Hence, both the quantities f and (4 a + b)f run through the reduced residual system modulo 32. Thus,

32

1 /n2s I V^'I if k 1 kl k 21 k 4i\

1 Kk(32) — 1 1 < ^ + E I cos If = ^ + Kcosk 9 + cosk 21 + cosk-9). (9) 2 =1 3 2

Denote the right-hand-side of (9) by g(k). Since g(k) is decreasing function of k then for any k ^ 13 we have

I Kk (32) — 1 I = g( k) ^ ^(13) = 0.953621..., Kk (32) > 0.0463788 ... > 1. The last inequality together with (6)-(8) yields:

Kk(32) > 0.0463788... > ^ for any k ^ 8. Finally, let n ^ 3. Then, for k ^ 10, Lemma 2.4 implies the inequality

18 ■ 3-k/2

Ak (3') <

=3

EAk (3r)

1 — 31- k/2

Setting h(k) for right-hand-side and using (9) we find

I Kk(q) — 1 I < g(k) + h(k) for any n ^ 3 and q = 3n. Since h(k) ^ h(13) < 0.0142895 ... fa any k ^ 13, we get

I Kk (q) — 1 I < ^(13) + h(13) < 0.967911..., Kk (q) > 0.0320892 ... > -1.

32

For 10 ^ k ^ 12, the inequalities (7), (8) together with the bound

Kk(3n) > Kk(32) — h(k)

imply

Kio(3n) > 0.0966797... — 0.075 > 0.021 > —,

50

Kn(3n) > 0.0966797... — 0.0430737 > 0.053606 > -1,

19

Ki2(3n) > 0.171387... — 0.0247934 > 0.1465936 > 1.

= 8, 9

Kk(3га) = Kfc(32) + ^ Ak(3r) ^ Kfc(32)

54-3

-k

r=4

1 — 31-k/2 '

The direct calculation with all triples (a,b,m), 1 ^ a,b,m ^ 33, (ab, 3) = 1, a + b ^ 0 (mod 3) shows that

Kk (32) > 0.0347222 ..., k = 8, 9.

Therefore,

Kg (3n) > 0.0347222... — 0.00854701 > 0.026175 > —,

«i / 39

Kg (3n) > 0.0347222... — 0.00280343 > 0.031918 > -1.

32

□

Corollary. For any k ^ 8; the exceptional set Qk(3n) consists precisely of the triples pointed in Corollary of Lemma 3.1.

Now we proceed to study the values of Kk(a, b, m; 3n) for 3 ^ k ^ ^d n ^ 2. In view of Corollary to Lemma 3.1, we will assume that a + b ^ 0 (mod 3) or, that is the same, a = b (mod 3)

Lemma 3.3. Suppose that 3 ^ k ^ 7. Then the set Qk(32) contains 18(8 — k) triples (a,b,m), a = b (mod 3) (ab, 3) = 1 listed in (10)-(13).

Proof. Indeed, let Q = = Qb,a be the set of values of the function g(x) = ax + bx on Zg. The direct computation shows that

Q = {1, 8} for the pairs (a; b) G £1 = {(1 Q = {2, 7} for the pairs (a; b) G £2, £2 = {(1 Q = {4, 5} for the pairs (a; b) G £3, £3 = {(1

7), (7 1), (2 4), (4

1), (2 5), (5 1), (2

8), (8 2), (4 2), (5

2), (4

7), (7

8), (8

4), (5

4), (8

5), (7

5)}, 8)}, 7)}.

Also, the direct computation shows that the sets kQ = p + ... + ^ do not coincide with complete

k

residual system Z9 for 3 ^ k ^ 7. Thus, the sets 3^ have the forms

Zg \ {0, 2, 4, 5, 7}, Zg \ {0,1, 4, 5, 8}, Zg \ {0,1, 2, 7, 8} respectively, the sets 4^ have the form

Zg \ {1, 3, 6, 8}, Zg \ {2, 3, 6, 7}, Zg \ {3, 4, 5, 6}, respectively, the sets 5^ have the form

Zg \ {0, 2, 7}, Zg \ {0, 4, 5}, Zg \ {0,1, 8}, respectively, the sets 6^ have the form

Zg \ {1, 8}, Zg \ {2, 7}, Zg \ {4, 5}, respectively, and, finally, all the sets 7^ coincide with Zg \ {0}. Hence, k3(32) = 0 for the triples

(a, b, 0), ( a, b, 2), ( a, b, 4), (a, b, 5), (a, b, 7) ( a; b) G £1, (a,b, 0), ( a,b, 1), ( a,b, 4), (a,b, 5), (a,b, 8) ( a; b) G £2, (a,b, 0), ( a, b, 1), ( a, b, 2), (a,b, 7), (a,b, 8) ( a; b) G £3,

k4(32) = 0 for the triples

(a,b, 1), (a,b, 3), (a,b, 6), ( a,b, 8), (a;b) e , (a,b, 2), (a,b, 3), (a,b, 6), ( a,b, 7), (a;b) e £2, (a,b, 3), (a,b, 4), (a,b, 5), ( a,b, 6), (a; 6) e £3,

(11)

k5(32) = 0 for the triples

( a,b, 0), ( a,b, 2), (a, b, 7), ( a; 6) e£b ( a,b, 0), ( a,b, 4), (a,b, 5), ( a; 6) e £2, ( a, 6,0), ( a, b, 1), (a, b, 8), ( a; 6) e £3,

(12)

k6(32) = 0 for the triples

(a,b, 1), (a,b, 8), (a;b) e £1,

(a,b, 2), (a,b, 7), (a; 6) e £2, (13)

(a,b, 4), (a,b, 5), (a; 6) e £3,

and, finally, k7(32) = 0 for the triples (a, b, 0), where ( a; 6) e £1 U £2 U £3. It is not difficult to check that the number of such triples coincides with 18(8 — k) in each case. This proves the lemma. □

Corollary. For any k, 3 ^ k ^ 7, we have

□

Suppose now that n ^ 3 and the triple (a, b, m) with the conditions 1 ^ a, b,m ^ 3n, (ab, 3) = 1 is congruent with some exceptional triple (a',b',m') e Qk(32) modulo 32. Then the triple (a,b,m)

Q k (3 n)

is also solvable, and that is impossible (here xx = 1 (mod 3n), xx* = 1 (mod 32), so x = x* (mod 32) Q k(3 n)

Lemma 3.4. Suppose that n ^ 3 3 ^ k ^ 7. Then there exists an absolute constant c4 > 0 such that the inequality

holds for any triple ( a,b,m), 1 ^ a,b,m ^ 3n; (ab, 3) = 1; that does not coincide with some

Q k(32) 32

Proof. By Lemma 2.4,

Qk (32) I = 18(14 — k).

g(x1) + ... + g(xk) = m (mod 3n), g(x) = ax + bx,

is solvable, then the congruence

g'(x1) + ... + g'(xk) = m (mod 32), g'(x) = a'x* + b'x,

Kk (3n) = Kk ( a,b,m, 3n) > C4

Kk(3n) — Kk(33) I <

54•3-k

and therefore Kk(3n) > Kk(33)

54 3- k

1- 31-k/2

1-31-k/2'

In particular,

К6(3га) > к6(33) - -2, Х7(3га) > К7(33) - 27 + л/3 - "

332

> К7(33) - 37.

(14)

The direct calculation with all triples ( a, b,m), 1 ^ a,b,m ^ 33, (ab, 3) = 1, that are not congruent with triples from Qk(3k) modulo 32 (k = 6, 7) shows that the least values of k6(33) and k7(33) are 25 13

equal to —— and ——. Thus, by (14) we conclude that

192

192

K6(3ra) > - 1 = 3, Kz(3ra) > -13 - - > -.

192 12 64'

192 37 25

The calculations also shows that

K5(3") > -7

K4(3") > -3

for n = 3, for n = 3, for n = 3, 4, 5

(15)

(16) (17)

for any triple (a,b,m), 1 ^ a,b,m ^ 3n, (ab, 3) = 1 that does not coincide with some triple from the set Qk(32) modulo 32 (3 ^ k ^ 5). At the same time, the inequalities of Lemmas 2.7, 2.6 and 2.5 imply that

35

1*5(3») - K5(33)| < —

3

|К4(3») - K4(33)| < -о

i r i 9

| К3(3») - K3(35) | < 16

for n ^ 4, for n ^ 4, for n > 6.

(18)

(19)

(20)

Thus the relations (15)-(20) yield:

35

1

35

1

K5(3ra) > K5(33) - 4!6 > 4 - 416 > 7

K4(3») > K4(33) - 3 > 176 - 3 > 16

K3(3») ^ K5(35) - 3 ^ 3 - 9 > 3

16 4 16

16

for n ^ 4, for n ^ 4, for n > 6.

Taking c4 = —, we arrive at the assertion of the lemma. □ 37

Corollary. For any 3 ^ k ^ 7, n ^ 2 one has

|Q k(3n)| = 33(n-2) ■ 18(14 — k) = 2 ■ 33n-4(14 — k).

This assertion finishes the description of the quantities (3ra), k ^ 3, n ^ 1.

СПИСОК ЦИТИРОВАННОЙ ЛИТЕРАТУРЫ

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Получено 23.01.2019 г. Принято в печать 20.03.2020 г.