On some diophantine inequalities involving primes

(3 6 - H )  
1
| 
J  
| < cn B N (log N) 
8 
J  | S ( b xa) S ( b2a) | da
o
and, from Holder’s inequality, the integral on the right is a t m ost [}VxW 2f  where
v 
i
W j = f  | S(bja) ¡2 da, 
(j == 1, 2).
. 
0
Now W j is m erely th e num ber of primes and hence, on noting th a t I I  > 120, we 
obtain
( 3 6 - / / )
H
| / 1 ^ cu B N 2{log N) 
* 
^
log TV) 
i<>.
1
1
 
1 
On th e other hand, since ~ H  — 3 > - ^ H  and, by hypothesis, @(m) >
(14)
implies th a t th e second term on the right of (10) is less th an half the first term . Thus we
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B a k e r , On some diophantine inequalities involving primes

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have
| 
I  | > — (6 
B 2)~2N
(log 
and it follows, on using (14) again, th a t J / [ > | 
|. This proves Lemma 6.
Lemma 7. Either 8 (1 ) > y or 8 (2 ) > y *
Proof. For each fixed /re, Cq(m) is a m ultiplicative function of q. F urther, it is 
easily verified th a t for each prime 
p and non-negative integer I, C^{m) is given by 
p l~1(p — 1), — p i_1 or 0 according as 
I 5
^ 
k, + 1 or 
>
+ 1, where k is th e 
exponent to which p  divides m. Thus
8 (w) = n
¿ [ h c A b ^ C
p  
/ = 0 
0 = 1  
v
J 
p  
p  prim e
and on recalling our supposition th a t (b1, b2, b3) = 1 we obtain
<£(m) = I I X(p),
v
where
X(p) = 1 +
{ J 
Cp{b,) 
1
C p(m) ( 
) ) - 3.
Now x(p) is given by 1 — (p — l ) -2, 1 
+ {p — I ) -1, 0 or 1 +
— l ) -3 according as p  
divides exactly one, two, three or none of 
/re. Thus
<3(1) > il {1 — (/> — I)-2} > 77 (1 
— P ~ 2)
>
i
p > 2 
p
*
except in th e case th a t ju st one of 
b2, 
b2, b2 is even. However if ju st one of bu b2, b3 is 
1
even then 8 (2 ) > -¡r- and th is proves the lemma.
JL
Lemma 8. For all real x we have 
v
CO
f e(ocx) {sin (7ia)l(7za)}2dcc =  m ax (0, 1 — | x  |).
— QO
Proof. See D avenport and Heilbronn [2], Lem ma 4, p. 188.
Lem m a 9. Suppose that a, /?, rh £ are real numbers with rj irrational and 0 < /? — a < 1.
I
Let d =-£-(/? — <*) and let N be any integer exceeding the maximum of d~2 || krj ||_1 for 
k = 1, 2 ,. 
[5~2]. Then the number of integer triplets i, m, n with 2 ^ m, n 5S N satisfying
—... 
. 
ec 
mrj + rc£— I < P
exceeds d N 2.
Proof. We proceed by the well known m ethod, originated by Weyl [6], for investigat­
ing th e distribution of sequences modulo 1 5).
B a k e r , On some diophantine inequalities involving primes 
173
tT
5) 
In connexion with Lemma 9, Professor D avenport has pointed out the paper of P . Erdos and P . T urd n , 
On a problem in the theory of uniform distribution, Nederl. Akad. W etensch. 51 (1948), (I) 1146— 1154, (II) 
1262— 1269 = Indag. M a th /10 (1948), (I) 370—378, (II) 406—413. On p. 1150 ( = p. 374) of this paper there are 
stated two explicit forms of WeyTs theorem (one proved in the paper as Theorem III and the other due to Van 
der Corput and Koksma) from either of which Lemma 9 can be deduced directly, apart from an undetermined 
numerical constant. It is convenient, however, to have precise estim ates for all constants appearing in the result 
and we have therefore retained the above proof.

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For any <%, /3 with 0 < /3 — oc < 1 let F(oc, /3, N) be the num ber of integer triplets
1
I, m, n w ith the properties required by the lemma. We p ut xp(x) = x  — [x ] — ^  and 
note th a t if x  is not an integer then
® sin (2n k x )
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B a k e r, On some diophanline inequalities involving primes
(15) 
y>(x)
i
Since [x — a] — [x — /3] is 1 or 0 according as an integer I does or does not exist 
such th a t oc 
x  — I < /3, we clearly have
F{x, 0, N) = {P — x) {N —
l)2 + i J
{y,(mi, + nZ — P) — y>(mri + nC — *)}.
m —
2 n = 2
Then w riting !F(:r) for
8
d~l f y)(x + t)dt 
0
we obtain
(16) 
r 1/ F (* + <5 — i ,£ — /, 
N) dt = (p —
X
— d ) ( N —  1)2 + £ /— F,
where
A c
U = 2
2 W { m r t
+ reC — £),
and F is defined sim ilarly w ith /3 replaced by a +
Furth er, from (15) we see th a t
(17) 
U — ^ - n ~ 2{ 2 +
u{k)\,
*