Fractional parts of polynomials over the primes. II
Roger Baker

TL;DR
This paper improves the understanding of how closely quadratic polynomials with irrational leading coefficients evaluated at prime numbers approximate integers, demonstrating that the fractional parts are very small for infinitely many primes.
Contribution
It provides a 14% improvement in the exponent measuring the proximity of polynomial values to integers at prime arguments compared to previous results.
Findings
Fractional parts of quadratic polynomials at primes are very small infinitely often.
Achieved a 14% enhancement over previous bounds.
Results contribute to understanding polynomial behavior over primes.
Abstract
We consider the distance to the nearest integer of f(p), where f is a quadratic polynomial with irrational leading coefficient. This distance is very small as a function of p, for infinitely many primes p. We give a 14% improvement in the exponent that measures the distance, compared with the most recent result in the literature.
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Taxonomy
TopicsAnalytic Number Theory Research · Meromorphic and Entire Functions · Algebraic Geometry and Number Theory
Fractional parts of polynomials over the primes. II
Roger Baker*†*
Department of Mathematics
Brigham Young University
Provo, UT 84602, U.S.A
To Glyn Harman on his sixtieth birthday.
Abstract.
Let denote distance from the integers. Let , , be real numbers with irrational. We show that the inequality
[TABLE]
has infinitely many solutions in primes , sharpening a result due to Harman (1996) in the case and Baker (2017) in the general case.
Key words and phrases:
Fractional parts of polynomials, exponential sums over primes, generalized Vaughan identity, Harman sieve, linear sieve.
2010 Mathematics Subject Classification:
Primary 10J54, Secondary 11L20, 11N36
*†*Research supported in part by Collaboration Grant 412557 from the Simons Foundation
1. Introduction
Let be a polynomial of degree with irrational leading coefficient. Inequalities of the form
[TABLE]
for infinitely many primes were studied by Vinogradov [11]; see [1] for the strongest available results. The present paper gives a new result for .
Theorem 1**.**
Let ; then (1.1) holds for infinitely many primes .
The values , , were given by Ghosh [6], Baker and Harman [2], and Harman [7] in the case ; Baker [1] extended the last result to general .
As in [1], [2], [7] we use the Harman sieve. We make progress in the present paper by giving new bounds for sums of the shape
[TABLE]
where is the approximating polynomial to in [1]. Type I sums (in which ) are treated in Section 2, and general (Type II) sums in Section 3. In the Type I case, is restricted to convolutions of shorter sequences; a lemma of Birch and Davenport [4] on Diophantine approximation plays a key role. For Type II sums, a subsidiary task is the study of the average behavior, as varies, of the number of solutions of
[TABLE]
One sum that eludes the Harman sieve takes the form
[TABLE]
(where , and ). We bound it above using the form of the linear sieve given by Iwaniec [10]. This sum is treated in Section 5; the sums accessible via the Harman sieve are in Section 4. Section 6 contains the sieve decomposition of (defined below), and the calculations leading to Theorem 1. Integrals that appear here and in earlier drafts were calculated by Andreas Weingartner; thanks, Andreas, for your generosity.
The following notations will be used:
.
.
indicator function of .
absolute constant, not the same at each occurrence.
real number with , not the same at each occurrence.
sufficiently small positive number; .
indicate implied constants that may depend on .
indicates .
.
fraction in lowest terms with , with sufficiently large; .
We choose so that
[TABLE]
and write
[TABLE]
Here will ultimately be ; earlier in the paper we restrict somewhat less. We do suppose , and write
[TABLE]
We reserve the symbols for prime numbers.
Let and
[TABLE]
We write for an arbitrary subinterval of .
2. Type I sums
The object of this section is to prove
Theorem 2**.**
Let . Let ,
[TABLE]
Let
[TABLE]
Then
[TABLE]
We require several lemmas.
Lemma 1**.**
Let . Let be the set of with and
[TABLE]
There are a set and positive numbers , with the following properties.
- (i)
We have
[TABLE] 2. (ii)
for , we have
[TABLE]
for some in in , and
[TABLE] 3. (iii)
* satisfies*
[TABLE]
Proof.
This can readily be extracted from the proof of [1, Lemma 8], with in place of . ∎
Lemma 2**.**
Let be a real number and suppose there exist distinct integer pairs satisfying
[TABLE]
where . Then all integer pairs , satisfying (2.6) have the same ratio .
Proof.
Birch and Davenport[4]. ∎
Lemma 3**.**
Suppose that , , , . The number of solutions , with , of the inequality
[TABLE]
is
[TABLE]
Proof.
[1, Lemma 3]. ∎
Proof of Theorem 2.
In the notation of Lemma 1, with , let be the set of pairs for which , , . As in [1], proof of Lemma 8, it suffices to prove that
[TABLE]
In view of the Type I result obtained in [1, Lemma 8], we may suppose that
[TABLE]
We note that
[TABLE]
as a consequence of Lemma 3; indeed
[TABLE]
since for sufficiently small.
Suppose for a moment that
[TABLE]
then (2.2) gives (with a divisor argument)
[TABLE]
(from (2.8),(2.9)), giving (2.7). So we may suppose that
[TABLE]
It now follows from (2.5) that
[TABLE]
Now let
[TABLE]
and for 1, let
[TABLE]
Then
[TABLE]
We choose , so that
[TABLE]
Suppose for a moment that
[TABLE]
Then arguing as above,
[TABLE]
Thus we may suppose that
[TABLE]
For the next stage of the argument, let be a fixed integer in . We apply Lemma 2, taking
[TABLE]
since (by (2.3) and the definition of )
[TABLE]
for every in . By a divisor argument, the number of distinct as varies over is . Thus in the notation of Lemma 2,
[TABLE]
To see this,
[TABLE]
from the hypothesis of the theorem.
Accordingly, all with can be written in the form
[TABLE]
for a certain , independent of .
We record a lower bound for that does not contain . From (2.12), (2.10), we have
[TABLE]
We now select a divisor of such that the set
[TABLE]
satisfies
[TABLE]
For each in , we have
[TABLE]
It is convenient to write where is square-free, and define . Then
[TABLE]
This leads to the upper bound
[TABLE]
from which we infer, using (2.15), that
[TABLE]
Now we re-examine our rational approximation
[TABLE]
(see (2.3) and the definition of ). We observe that
[TABLE]
for , using (2.16).
Let be the set of natural numbers , , , . The number of possibilities for here is , since is a divisor of . Hence
[TABLE]
All the integers occurring in (2.17) are in . Hence
[TABLE]
on bounding the number of in with via Lemma 3. Combining (2.18), (2.19), and recalling (2.11),
[TABLE]
We now use the lower bound (2.14) for and obtain
[TABLE]
Now (2.7) follows on applying the bounds for and in the hypothesis of Theorem 2. This completes the proof of Theorem 2. ∎
3. Type II sums
Lemma 4**.**
Let . For , let denote the number of quadruples with , such that
[TABLE]
Proof.
For (3.2), fix and , then the equation
[TABLE]
(with , ) determines , up to possibilities.
For (3.3), we observe that
[TABLE]
is the number of tuples , , , , , , , with , and
[TABLE]
This may be expressed as an integral:
[TABLE]
by the Cauchy-Schwarz inequality, where
[TABLE]
Now is the number of solution of
[TABLE]
with , , and .
We first consider , the number of solutions of (3.5) with
[TABLE]
If (3.5) and (3.6) hold, then . There are possibilities for , , , and for each of these, at most possibilities for , , . Thus
[TABLE]
Now consider , the number of solutions of (3.5) for which (3.6) is violated. There are possibilities for , , , . For each of these, there are possibilities for , and , hence possibilities for , , . Thus
[TABLE]
Now(3.3) now follows on combining(3.4) and (3.7). ∎
Theorem 3**.**
For , and , , , , we have
[TABLE]
Proof.
Just as in [1, proof of Lemma 9] we need only show that
[TABLE]
Again arguing as in that proof,
[TABLE]
where , and
[TABLE]
The contribution to the right-hand side of (3.9) from those with
[TABLE]
is
[TABLE]
since . The contribution from those with
[TABLE]
It remains to consider , the set of in with
[TABLE]
and
[TABLE]
We apply [1, Lemma 5] with , , . We require
[TABLE]
which holds since
[TABLE]
Thus for each there exists a natural number ,
[TABLE]
with , , ,
[TABLE]
Now and, writing ,
[TABLE]
Thus we can appeal to [1, Lemma 7] with and replaced by 1. Let
[TABLE]
We obtain
[TABLE]
Now is of smaller order than , so that
[TABLE]
Moreover, by standard bounds, the left-hand side of (3.12) is
[TABLE]
We now use a standard splitting-up argument to choose a subset of such that
[TABLE]
where , with
[TABLE]
and moreover
[TABLE]
while
[TABLE]
Compare e.g. the argument in [1, proof of Lemma 8]. In order to obtain (3.8) it remains to show that
[TABLE]
Using (3.11)–(3.14), we find that
[TABLE]
and we must show that
[TABLE]
For each in there is an with
[TABLE]
where
[TABLE]
Clearly
[TABLE]
Given , let , , be the divisors of with . Each element of satisfies
[TABLE]
for some and some , . Let
[TABLE]
Then
[TABLE]
while
[TABLE]
in the notation of Lemma 4. Applying Cauchy’s inequality,
[TABLE]
Alternatively, (3.2) yields
[TABLE]
We now find that, depending in the value of , either (3.18) or (3.19) yields the desired bound (3.15). Suppose first that
[TABLE]
In view of (3.16), (3.18), we need to verify the four bounds
[TABLE]
First of all, (3.21) holds since
[TABLE]
Next, (3.22) holds since
[TABLE]
Next, (3.23) holds since
[TABLE]
Finally, (3.24) holds since
[TABLE]
from (3.20).
Now suppose that
[TABLE]
we employ (3.16) and (3.19). We need to verify the bounds
[TABLE]
First of all, (3.26) holds since
[TABLE]
from (3.25). Next, (3.27) holds since
[TABLE]
Next, (3.28) holds since
[TABLE]
Finally, (3.29) holds since
[TABLE]
In order to complete the proof, we show that the ranges of in (3.20) and (3.25) overlap. We have
[TABLE]
that is
[TABLE]
since . This completes the proof of (3.15), and Theorem 3 follows. ∎
4. Asymptotic formulae via Harman sieve and generalized Vaughan identity
In the present section and the next, we suppose that satisfies
[TABLE]
We write , , and
[TABLE]
For a finite set , let
[TABLE]
As in [1], our claim in Theorem 1 is a corollary of the lower bound
[TABLE]
for .
We introduce some ‘comparison’ results for the pair and , and similar pairs, that will be needed in Section 6. First of all, we have
[TABLE]
whenever is a positive constant and some subproduct of
satisfies
[TABLE]
This is a consequence of Theorem 2; compare the discussion in [8, Sections 3.2 and 3.5]. Additional inequalities such as ‘’ may be included in the summation in (4.3) without affecting its validity, as explained in [8, Section 3.2].
The following lemma is essentially the special case , of [3, Lemma 14], and is a variant of [8, Theorem 3.1].
Lemma 5**.**
Let be a complex function with support in , . Let . Let
[TABLE]
Suppose that, for some we have, (for any coefficients , , and ,
[TABLE]
Let be complex numbers with , for . Then
[TABLE]
We can deduce the following ‘bilinear’ lemma.
Lemma 6**.**
Let , , , be as in Lemma 5. Suppose that we have the hypothesis (4.6) and in addition, for some ,
[TABLE]
for any , with , . Then for any , with , , for , we have
[TABLE]
Proof.
We apply Lemma 4 with replaced by ,
[TABLE]
so that is replaced by
[TABLE]
From (4.8), (4.9) the hypotheses of Lemma 5 are satisfied with replaced by : for example,
[TABLE]
(we may group the product as a single variable and apply (4.6)). The conclusion (4.7) with replaced with gives the desired bound (4.9). ∎
We now apply Lemma 6 with
[TABLE]
, , where and the non-negative number satisfies
[TABLE]
Lemma 7**.**
Suppose that (4.10)–(4.12) hold. Then
[TABLE]
whenever , for , and .
Proof.
We take . The hypothesis (4.8) is a consequence of Theorem 2 because of (4.10)–(4.12). The hypothesis (4.6) is a consequence of Theorem 2. Now the conclusion (4.9) may be written in the form (4.13). ∎
Lemma 8**.**
Let , . Let with , . There is a set with .
Proof.
Suppose that no such exists. Now suppose first that . Since we can prove successively that are in . This is absurd.
Thus we must have . But now , and since . This is absurd. ∎
Lemma 9**.**
*Let be a complex function on . The sum
may be decomposed into at most sums of the form*
[TABLE]
where , and if .
Proof.
This is a case of Heath-Brown’s ‘generalized Vaughan identity’ [9]. ∎
Lemma 10**.**
Let
[TABLE]
We have
[TABLE]
Proof.
Arguing as in the proof of (4.3), it will suffice to show that
[TABLE]
for . By a partial summation argument, it suffices to obtain
[TABLE]
Applying Lemma 9, we need only show that
[TABLE]
whenever and for . Thus . We write .
We may assume in view of Theorem 3 that no subproduct of satisfies
[TABLE]
We now reorder as with . Let , so that and
[TABLE]
We divide the argument into two cases.
Case 1**.**
We have . Thus and (after a partial summation if necessary) we can apply Theorem 2 to the sum in (4.15) with
[TABLE]
We verify the hypotheses of Theorem 2. First,
[TABLE]
since
[TABLE]
Next,
[TABLE]
and
[TABLE]
from the definition of . Now (4.15) follows from Theorem 2.
Case 2**.**
We have . By Lemma 8 and the absence of a product satisfying (4.17), there is a subsum such that
[TABLE]
We are now in a position to apply Lemma 6 with , ,
[TABLE]
We need to verify (4.6), (4.8). Clearly (4.8) is a consequence of Theorem 3. As for (4.8), we need to verify the hypotheses of Theorem 2 with , . Suppose first that . Then ,
[TABLE]
Now suppose that . Then ,
[TABLE]
[TABLE]
and
[TABLE]
Thus Theorem 2 yields the desired estimate (4.8). Now the lemma follows from Lemma 6.∎
There is a short interval in which we can use Lemma 9 directly to obtain a conclusion stronger than (4.15).
Lemma 11**.**
We have
[TABLE]
whenever
[TABLE]
Proof.
In this range of we have
[TABLE]
similarly with in place of . We apply Lemma 9 to decompose the sum over (with in place of ). Clearly it will be enough to show that, for ,
[TABLE]
where and for . As in the preceding proof we may suppose that no subproduct of satisfies (4.17). Writing , we have
[TABLE]
Thus it is clear by a ‘reflection’ argument that no such can satisfy
[TABLE]
Moreover,
[TABLE]
It follows that
[TABLE]
There cannot be two indices with , since
[TABLE]
Hence there is a with
[TABLE]
We are now in a position to apply Theorem 2 with
[TABLE]
We make the usual verification:
[TABLE]
Thus Theorem 2 yields (4.18). This completes the proof of Lemma 11. ∎
5. Application of the linear sieve
In order to obtain an upper bound for
[TABLE]
whenever
[TABLE]
we apply Theorem 4 of Iwaniec [10], which we state in a form sufficient for our needs. The quantity estimated will actually exceed that in (5.1), which will be exploited in Section 6.
Let be a set of integers in . Fix an approximation to and write
[TABLE]
Let be the upper bound function for the linear sieve [10, p. 309]. In the following lemma, let . Let
[TABLE]
Lemma 12**.**
With the above notations, we have
[TABLE]
with some coefficients and bounded by in absolute value. Here
[TABLE]
In our application, we shall take
[TABLE]
for any , and
[TABLE]
We write . We shall take and . It is easily verified that , so that
[TABLE]
We apply Lemma 11 and sum over , obtaining (on noting that )
[TABLE]
Here
[TABLE]
We shall show that
[TABLE]
the proof that is similar but simpler.
Reducing the task of bounding to estimating exponential sums as in previous sections, it suffices to prove that for , and a fixed , we have
[TABLE]
This is a consequence of Theorem 2, grouping the variables as and ,
[TABLE]
We make the usual verifications:
[TABLE]
The desired bound (5.5), and the same bound for , now follows. Now (5.3), (5.4) yield
[TABLE]
6. The sieve decomposition
For the final stage of our work we take so that . In this section, each that occurs takes the form
[TABLE]
where the asterisk indicates a restriction of and to certain subsets of depending on ; is obtained from on replacing by . For some of these values of we write
[TABLE]
where is defined by the following additional condition of summation within : a subproduct of satisfies
[TABLE]
We split up as in the same way. As noted in Section 5,
[TABLE]
whenever (6.1) is used.
Conversion of sums into integrals, with an acceptable error, in the following is along familiar lines (see [8]). Concerning Buchstab’s function , we note that
[TABLE]
provided that , and that
[TABLE]
see Cheer and Goldston [5].
Let . Using Buchstab’s identity and writing , we have
[TABLE]
where
[TABLE]
with ,, , , and .
Recalling (4.3) and Lemmas 7 (with ) and 11, we have
[TABLE]
for . For , we apply Buchstab’s identity three further times to obtain
[TABLE]
where
[TABLE]
We can apply Lemma 7 to , , to obtain (6.7): for example, in ,
[TABLE]
We treat via (6.3).
For we apply Buchstab’s identity once. Iterating once more for part of the sum over , , this gives
[TABLE]
where
[TABLE]
We have (6.7) for , since Lemma 9 is applicable. For example, for we have
[TABLE]
We have (6.7) also for , this time using (4.3), since in . For , we use the lower bound (6.3).
We also apply Buchstab once more to ,
[TABLE]
where
[TABLE]
satisfies (6.7) by Lemma 10, and
[TABLE]
is bounded below as in (6.3).
For , we proceed differently. We have
[TABLE]
where
[TABLE]
is treated as in (6.3).
By (5.7),
[TABLE]
Moreover,
[TABLE]
Combining (6.11)–(6.13) and (6.3) with ,
[TABLE]
Our sieve decomposition, obtained by combining (6.6), (6.8), (6.9) and (6.10), is
[TABLE]
and also holds if is replaced by . Combining all applications of (6.7) and (6.3) with (6.14), we end up with
[TABLE]
Our next task is to evaluate , , , and with sufficient accuracy. Using we find that in , implies . With a little thought we find that
[TABLE]
It is simplest to replace the factor by using (6.4). Carrying out the and integrations (and using ) leads to
[TABLE]
For , we write the numbers that appear in as
[TABLE]
where , , ,
[TABLE]
We now consider . First we treat the contribution from . Thus . We cannot have , since
[TABLE]
If we consider , we have and obtain
[TABLE]
Let be the contribution to from . We have , , by an argument used above. We cannot have since
[TABLE]
If , we must have (use (6.20) again). If , we must have similarly; however, , since
[TABLE]
The three remaining cases lead to
[TABLE]
where
[TABLE]
(The integral was bounded above by an integral in , , , using , before carrying out the integration.)
In , we cannot have , since then
[TABLE]
which is absurd. Thus ; we cannot have or , or , since
[TABLE]
For , we have since
[TABLE]
Thus
[TABLE]
where
[TABLE]
In , we write the numbers that appear as as
[TABLE]
where , , , , ,
[TABLE]
We observe that
[TABLE]
Thus we cannot have or , or , . We cannot have , since
[TABLE]
Since , the only remaining possibility is , , so that
[TABLE]
where
[TABLE]
Combining this with (6.14) we obtain
[TABLE]
with
[TABLE]
(we obtain this using (6.5)). Since , we only need to add up our integrals to obtain Theorem 1 from (6.15)–(6.28):
[TABLE]
The reference list from the paper itself. Each links out to its DOI / PubMed record.
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- 2[2] R. C. Baker and G. Harman, On the distribution of α p k 𝛼 superscript 𝑝 𝑘 \alpha p^{k} modulo one, Mathematika 48 (1991), 170–184.
- 3[3] R. C. Baker and A. Weingartner, A ternary Diophantine inequality over primes, Acta Arith. 162 (2014), 159–196.
- 4[4] B. J. Birch and H. Davenport, On a theorem of Davenport and Heilbronn, Acta Math. 100 (1958), 259–279.
- 5[5] A. Y. Cheer and D. A. Goldston, A differential-delay equation arising from the sieve of Eratosthenes, Math. Comp. 55 (1990), 129–141.
- 6[6] A. Ghosh, The distribution of α p 2 𝛼 superscript 𝑝 2 \alpha p^{2} modulo 1, Proc. London Math Soc. 42 (1981), 252–269.
- 7[7] G. Harman, On the distribution of α p 𝛼 𝑝 \alpha p modulo one II, Proc. London Math. Soc. 72 (1996), 241–260.
- 8[8] G. Harman, Prime-detecting Sieves , Princeton University Press, Princeton, N.J. 2007.
