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Article

A Diophantine Inequality Involving Mixed Powers of Primes with a Specific Type

by
Tatiana L. Todorova
1 and
Atanaska Georgieva
2,*
1
Faculty of Mathematics and Informatics, Sofia University “St. Kliment Ohridski”, 1000 Sofia, Bulgaria
2
Faculty of Mathematics and Informatics, University of Plovdiv, 4000 Plovdiv, Bulgaria
*
Author to whom correspondence should be addressed.
Mathematics 2025, 13(19), 3065; https://doi.org/10.3390/math13193065
Submission received: 14 August 2025 / Revised: 20 September 2025 / Accepted: 22 September 2025 / Published: 23 September 2025
(This article belongs to the Special Issue Recent Studies in Number Theory and Algebraic Geometry)

Abstract

Let λ 1 , λ 2 , λ 3 be nonzero real numbers, not all of the same sign; let λ 1 / λ 2 be irrational; and let η be any real number. We investigate the solvability of the inequality | λ 1 p 1 + λ 2 p 2 + λ 3 p 3 2 + η | < ( max p j ) 1 / 12 + θ , θ > 0 in the prime variables p 1 , p 2 , and p 3 . We require that p 1 + 2 and p 2 + 2 have no more than 20 prime factors, while p 3 + 2 has no more than 42 prime factors.

1. Introduction

In 1946, Davenport and Heilbronn [1] made a significant contribution by adapting the Hardy-Littlewood method. Their work provided valuable insight and helped to prove that if λ 1 , , λ s are nonzero real numbers, not all of the same sign, and not all in a rational ratio, then for every E > 0 the inequality
j = 1 s λ j x j k < E
has infinitely many solutions in natural numbers x j provided that s 2 k + 1 . Later, Davenport and Roth [2] proved that if k 12 , then s > C k log k suffices with a suitable absolute constant C. More recently, Schwarz [3] demonstrated that if either s 2 k + 1 or s 2 k 2 ( 2 log k + log log k + 5 2 ) 1 (for k 12 ), then the inequality
j = 1 s λ j p j k < E
has infinitely many solutions, with all p j being prime numbers.
In 1967, for the specific case k = 1 and s = 3 , Baker [4] advanced the field by demonstrating that when
λ i R , λ i 0 , i = 1 , 2 , 3
λ 1 , λ 2 , λ 3 not all of the same sign
λ 1 / λ 2 R Q
η R
there are infinitely many ordered triples of primes p 1 , p 2 , p 3 such that
| λ 1 p 1 + λ 2 p 2 + λ 3 p 3 + η | < E
with E = ( log m a x p j ) A , A > 0 . From now on, we will assume that the conditions in Equations (1)–(4) are met. The Baker’s approach laid the groundwork for further exploration in this area. Subsequently, the upper bound of the right-hand side of Equation (5) was sharpened by [5,6,7,8,9,10]
Ramachandra with E = e x p ( ( log p 1 p 2 p 3 ) 1 / 2 ) Vaughan with E = ( max p j ) 1 / 10 + θ Lau and Liu with E = ( max p j ) 1 / 9 + θ Baker and Harman with E = ( max p j ) 1 / 6 + θ Harman with E = ( max p j ) 1 / 5 + θ Matom ä ki with E = ( max p j ) 2 / 9 + θ
and here θ > 0 .
The best result to date belongs to K. Matomäki, but Baker and Harman [8] proved that under the generalized Riemann hypothesis, it is possible to reach E = ( max p j ) 1 / 4 + θ .
Recent research has concentrated mainly on cases involving a limited number of summands.
In 2012 Li Wei-ping, Wang Tian-ze [11] proved the existence of infinitely many triples of primes such that
| λ 1 p 1 + λ 2 p 2 + λ 3 p 3 k + η | < E ,
where k = 2 and E = ( max p j ) 1 / 40 + θ . Later, Langusco and Zaccagnini [12] proved Equation (6) with 1 < k < 4 / 3 and E = ( max p j ) 2 5 k + 3 10 + θ . They notice that the same arguments give Equation (6) with k = 2 and E = ( max p j ) 1 / 18 + θ . In [13], Gambini, Languasco and Zaccagnini proved that Equation (6) is fulfilled with 1 < k 3 and
E = ( max p j ) 3 2 k 6 k + θ if 1 < k 6 5 ( max p j ) 1 12 + θ if 6 5 < k 2 ( max p j ) 3 k 6 k + θ if 2 < k < 3 ( max p j ) 1 24 + θ if k = 3 .
A similar problem observed by Languasco and Zaccagnini [14] is the existence of infinitely many solutions in primes of the diophantine inequality | λ 1 p 1 + λ 2 p 2 2 + λ 3 p 3 k + η | < ( max p j ) 33 29 k 72 k + θ with 1 < k 33 29 . Later, Fu Linzhu, Liqun Hu, and Xuan Long [15] extended their result to k 3 .
Another type of Diophantine inequality is those with more summands and again with mixed powers. We mention some of them.
Wang and Yao [16] considered the inequality
| λ 1 p 1 + λ 2 p 2 2 + λ 3 p 3 2 + λ 4 p 4 2 + ϖ | max j p j 1 / 14 + θ
and proved that it has infinitely many solutions in prime variables p 1 , , p 4 . Under conditions λ 1 / λ 2 R Q , and if both λ 1 / λ 2 and λ 1 / λ 3 are algebraic, the author proved that the exponent can be replaced by 3 40 + θ . For the inequality
λ 1 p 1 + λ 2 p 2 2 + λ 3 p 3 3 + λ 4 p 4 4 + λ 5 p 5 k + η < E
with k = 5 , Zhu [17] proved that it has infinitely many solutions in the prime variables p 1 , , p 5 with E = ( max p j ) 19 756 + θ . Later, Mu [18] improved recent results by proving the solvability of this inequality with E = ( max p j ) 5 288 + θ for k = 4 and E = ( max p j ) 5 6 k 2 ( k + 1 ) for k 5 .
Alongside Diophantine inequalities involving mixed powers, inequalities that impose additional restrictions on the participating prime numbers are also of interest.
In 1991 Tolev [19] solved the diophantine inequality in Equation (5) in primes p 1 , p 2 , p 3 near to squares. More precisely he proved the existance of infinitely many triples p 1 , p 2 , p 3 satisfying (5) with E = ( max p j ) 1 / 8 + θ and such that
p 1 , p 2 , p 3 < ( max p j ) ( 1 8 τ ) / 26 log 5 ( max p j )
(as usual, α denotes the distance from α to the nearest integer). Later Dimitrov [20] proved (5) with E = ( max p j ) 37 c 38 26 ( max log p j ) 10 with Piatetski-Shapiro primes p i = [ n i c ] , n i N , i = 1 , 2 , 3 , 1 < c < 38 37 .
Another famous and still unsolved problem is the existence of infinitely many primes p such that p + 2 is also prime. Let P r be an integer with no more than r prime factors, counted with their multiplicities. In 1973, Chen [21] showed that there are infinitely many primes p with p + 2 = P 2 .
In 2015, Dimitrov and Todorova [22] mixed Vaughan and Chen problems, and for k = 1 and E = ( log max p j ) B they proved (5) with p 1 + 2 = P 8 , p 2 + 2 = P 8 , p 3 + 2 = P 8 . Dimitrov [23] improved this result with E = ( max p j ) 1 / 12 + θ and p 1 + 2 = P 28 , p 2 + 2 = P 28 , p 3 + 2 = P 28 . Later, the author [24] proved the existence of infinitely many triples of primes satisfying Equation (5) with E = ( max p j ) 1 / 18 + θ , p 1 + 2 = P 7 , p 2 + 2 = P 7 , and p 3 + 2 = P 7 . We refer to a hybrid theorem demonstrated by Dimitrov in [25], which establishes the inequality in Equation (5) that involves the primes p 1 , p 2 , p 3 , where p 3 is of the form x 2 + y 2 + 1 .
In the present paper, we will study the Diophantine inequality with mixed powers of primes, such that p i + 2 , i = 1 , 2 , 3 , are almost primes. More precisely, we prove the following theorem:
Theorem 1. 
If the conditions in Equations (1)–(4) are fulfilled and θ > 0 , then there are infinitely many ordered triples of primes p 1 , p 2 , p 3 with
| λ 1 p 1 + λ 2 p 2 + λ 3 p 3 2 + η | < ( max p j ) 1 / 12 + θ
and
p 1 + 2 = P 20 , p 2 + 2 = P 20 , p 3 + 2 = P 42 .

2. Notations

By p , p 1 , p 2 , p 3 , p 4 , we always denote primes. As usual, φ ( n ) and μ ( n ) denote, respectively, Euler’s function and Möbius’ function. Let ( m 1 , m 2 ) and [ m 1 , m 2 ] be the largest common divisor and the least common multiple of m 1 , m 2 . Instead of m n ( mod k ) , for simplicity, we write m n ( k ) . As usual, [ y ] denotes the integer part of y, e ( y ) = e 2 π i y .
Let χ be Dirichlet us character and let L ( s , χ ) be the corresponding L function. We will use the notation.
N ( β , T , χ ) = # ρ = σ + i γ | L ( ρ , χ ) = 0 , | γ | T , β σ
N ( T , χ ) = N ( 0 , T , χ )
N ( β , T ) = # ρ = σ + i γ | ζ ( ρ ) = 0 , | γ | T , β σ ,
where ζ ( s ) is the Riemann zeta function.
For D 1 and T 2 we denote
N ( β , T , D ) = d D χ * m o d d N ( β , T , χ ) ,
where ∗ means that the sum is taken over primitive characters modulo d. Also by
ψ ( y ) = n y Λ ( n ) and ψ ( y , a , d ) = n y n a ( d ) Λ ( n )
we denote Chebyshev’s functions,
E ( x , d , a ) = p x p a ( d ) log p x φ ( d )
and for a given character χ , we write
ψ ( y , χ ) = n y Λ ( n ) χ ( n ) .
We will write k K when K / 2 < k K . The letter ε denotes an arbitrarily small positive number, not the same in all appearances. For example, this convention enables us to write x ε log x x ε .

3. Auxiliary Results

In the proof of our Theorem, we will use a vector sieve, and we will need the following Lemma:
Lemma 1. 
Suppose that D R , D > 4 . There exists an arithmetical functions λ ± ( d ) (called Rosser’s functions of level D ) with the following properties:
1. 
For any positive integer d, we have
| λ ± ( d ) | 1 , λ ± ( d ) = 0 if d > D or μ ( d ) = 0 .
2. 
If n N then
d | n λ ( d ) d | n μ ( d ) d | n λ + ( d ) .
3. 
If z R is such that z 2 D and if
B ± ( D , z ) = d | P ( z ) λ ± ( d ) φ ( d ) , s = log D log z ,
then we have
B ( z ) B + ( D , z ) B ( z ) F ( s ) + O ( log D ) 1 3 ,
B ( z ) B ( D , z ) B ( z ) f ( s ) + O ( log D ) 1 3 ,
where F ( s ) and f ( s ) satisfy
F ( s ) = 2 e γ s if 0 < s 3 ; F ( s ) = 2 e γ s 1 + 2 s 1 log ( t 1 ) t d t if 3 s 5 ; f ( s ) = 2 e γ log ( s 1 ) s , if 2 s 4 ; ( s F ( s ) ) = f ( s 1 ) if s > 3 ; ( s f ( s ) ) = F ( s 1 ) if s > 2 .
Here, γ is Euler’s constant γ 0.577 .
Proof. 
See Greaves (Chapter 4, [26]) and [27]. □
From Merten’s formula, we observe that it follows:
B ( z ) = 2 C 2 e γ log z 1 + O 1 log z ,
where
C 2 = p 3 1 1 ( p 1 ) 2 0 , 66016 . . .
is the twin prime constant. From the properties of the functions of the linear sieve in Equations (16) and (17), it follows that
0 B ( D , z ) B ( z ) B + ( D , z ) .
We will utilize Bombieri-Vinogradov’s theorem.
Lemma 2. 
(Bombieri–Vinogradov) For any A > 0 the following inequality
q X 1 2 / ( log X ) A + 5 max y X max ( a , q ) = 1 | E ( y , q , a ) | X ( log X ) A .
is fulfilled.
Proof. 
See (ch.28, [28]). □
In the following Lemma, we provide explicit formulas for Chebyshev’s function ψ ( y ) and for the function ψ ( y , χ ) .
Lemma 3. 
Let 2 T y . Then
ψ ( y ) = y | γ | T y ρ ρ + O y log y T ,
where summation is taken over the non-trivial zeros ρ = σ + i γ of the Riemann zeta function such that | γ | < T .
If q > 1 and χ is a primitive character modulo q then
ψ ( y , χ ) = | γ | T y ρ ρ + | γ | 1 1 ρ + O y log ( q y ) T ,
where summation is taken over the non-trivial zeros ρ = σ + i γ of the Dirichlet us L-function L ( s , χ ) such that | γ | < T and | γ | < 1 , respectively.
Proof. 
See [28] §17 and [28] §19. □
We need the following two Lemmas regarding the zeros of the Dirichlet L-functions and the Riemann zeta function.
Lemma 4. 
For any ϱ > 0 there is a positive number C ( ϱ ) such that if χ is a quadratic character modulo q and σ is a real zero of L ( s , χ ) , then
σ < 1 C ( ϱ ) q ϱ .
Proof. 
See Corollary 11.15 of §11, [29]. □
Lemma 5. 
For | t | 3 the Riemann ζ function has no zeros in the region
σ > 1 1 53.989 ( log | t | ) 2 / 3 ( log log | t | ) 1 / 3
Proof. 
See Theorem 1 [30]. □
The next three Lemmas provide information about the density of the zeros of Dirichlet us L-functions and of Riemann’s ζ -function.
Lemma 6. 
Let χ be a primitive character modulo q and T 2 . Then
N ( T , χ ) T log ( q T ) ) .
Proof. 
See [28], §16. □
Lemma 7. 
Let t 2 and function N ( β , T , D ) be defined with (13). Then
N ( β , T , D ) ( D 2 T ) 3 ( 1 β ) 2 β log 9 ( D T ) if 1 2 β 4 5 ( D 2 T ) 2 ( 1 β ) β log 14 ( D T ) if 4 5 β 1 ( D 2 T ) ( 2 + ϵ ) ( 1 β ) if 4 5 β 1 .
Proof. 
See Theorem 12.2, §12, [31,32]. □
Lemma 8. 
Let T 2 and function N ( β , T ) be defined by Equation (12). Then
N ( β , T ) T log T if 0 β 1 2 T 3 ( 1 σ ) 2 σ ( log T ) 5 if 1 2 β 3 4 ; T 8 ( 1 σ ) 3 ( log T ) 5 if 0.52 β 1 .
Proof. 
See ch. 11 [33] and Theorem 1.1 [34]. □
Lemma 9. 
Suppose α R and α satisfy conditions
| α a q | < 1 q 2 , a Z , q N , ( a , q ) = 1 , q 1 .
Let ξ ( d ) be complex numbers defined for d D , D X 4 / 25 , and ξ ( d ) 1 . If
W ( X ) = d D ξ ( d ) p X p + 2 0 ( d ) e ( α p ) log p
then for any arbitrary small ε > 0 we have
W ( X ) X ε X q 1 / 2 + X 1 / 2 q 1 / 2 + X 2 / 3 D + X 5 / 6 .
Proof. 
This is Lemma 1 from [35]. □

4. Beginning of the Proof

Let 0 < λ 0 < 1 , ξ , δ , α , δ ˜ , α ˜ , be positive real numbers that we will specify later, but for now we will only assume the conditions.
ξ + 3 δ < 12 25 , ξ + δ < 3 8 , ξ + 3 δ ˜ < 6 25 , ξ + δ ˜ < 3 16 , 0 < α < δ / 2 , 0 < α ˜ < δ ˜ / 2 .
We define
Δ = X ξ 1
D = X δ , D ˜ = X δ ˜
E = X 1 / 12 + θ , θ > 0 is arbitrary small , H = 1000 log 2 X E
z = X α , z ˜ = X α ˜ ,
I ( β ) = λ 0 X X e ( β y ) d y
I ˜ ( β ) = λ 0 1 2 X 1 2 X 1 2 e ( β y 2 ) d y
P ( z ) = 2 < p < z p , B ( z ) = 2 < p < z 1 1 p 1 .
Consider the sum
Γ ( X ) = λ 0 X < p 1 , p 2 , p 3 X ( p i + 2 , P ( z ) ) = 1 , i = 1 , 2 ( p 3 + 2 , P ( z ˜ ) ) = 1 | λ 1 p 1 + λ 2 p 2 + λ 3 p 3 2 + η | < E log p 1 log p 2 log p 3 ,
with E given by Equation (26). If we can establish the inequality Γ ( X ) > 0 , then the inequality in Equation (8) would have a solution in the primes p 1 , p 2 , p 3 that satisfies the conditions ( p i + 2 , P ( z ) ) = 1 for i = 1 , 2 and ( p 3 + 2 , P ( z ˜ ) ) = 1 . If the number p i + 2 has multiplicity-counted prime factors represented by i , then from Equations (27) and (30), we can conclude that i < 1 α for i = 1 , 2 , and 3 < 1 α ˜ . This implies that p 1 + 2 and p 2 + 2 are almost primes of order 1 α , while p 3 + 2 is an almost prime of order 1 α ˜ .
To transform the sum Γ ( X ) we take a function υ such that
υ ( x ) = 1 f o r | x | 3 E / 4 0 < υ ( x ) < 1 f o r 3 E / 4 < | x | < E υ ( x ) = 0 f o r | x | E .
The function υ has derivatives of sufficiently large order, and its Fourier transform
Υ ( x ) = υ ( t ) e ( x t ) d t
satisfy
| Υ ( x ) | min 7 E 4 , 1 π | x | , 1 π | x | k 2 π | x | E / 8 k
for all k N . For the existence of such a function, see [36].
Using the function υ ( x ) , we get
Γ ( X ) Γ 0 ( X ) = λ 0 X < p 1 , p 2 , p 3 X ( p i + 2 , P ( z ) ) = 1 , i = 1 , 2 ( p 3 + 2 , P ( z ˜ ) ) = 1 υ ( λ 1 p 1 + λ 2 p 2 + λ 3 p 3 2 + η ) log p 1 log p 2 log p 3 .
Our goal is to demonstrate that for specific values of α and α ˜ (as large as possible), there exists a sequence X 1 , X 2 , such that Γ 0 ( X j ) > 0 . Then the number of prime solutions p i of Equation (8) in the interval ( λ 0 X j , X j ] with p i + 2 = P [ 1 / α ] for i = 1 , 2 and p 3 + 2 = P [ 1 / α ˜ ] is positive. This approach allows us to generate an infinite sequence of triples of primes p 1 , p 2 , p 3 that satisfy the desired properties.
Let Λ i = d | ( p i + 2 , P ( z ) ) μ ( d ) and Λ 3 = d | ( p 3 + 2 , P ( z ˜ ) ) μ ( d ) be the characteristic functions of primes p i , such that ( p i + 2 , P ( z ) ) = 1 for i = 1 , 2 and ( p 3 + 2 , P ( z ˜ ) ) = 1 , respectively. Then from Equation (34) follows that
Γ 0 ( X ) = λ 0 X < p i X i = 1 , 2 , 3 υ ( λ 1 p 1 + λ 2 p 2 + λ 3 p 3 + η ) Λ 1 Λ 2 Λ 3 log p 1 log p 2 log p 3 .
Let λ ± ( d ) and λ ˜ ± ( d ) represent the lower and upper bounds of Rosser’s weights at levels D and D ˜ , respectively (see Lemma 1). If
Λ i ± = d ( p i + 2 , P ( z ) ) λ ± ( d ) , i = 1 , 2 and Λ 3 ± = d ( p 3 + 2 , P ( z ˜ ) ) λ ˜ ± ( d )
then, from Lemma 1 we have Λ i Λ i Λ i + , i = 1 , 2 , 3 .
We will utilize the following simple inequality.
Λ 1 Λ 2 Λ 3 Λ 1 Λ 2 + Λ 3 + + Λ 1 + Λ 2 Λ 3 + + Λ 1 + Λ 2 + Λ 3 2 Λ 1 + Λ 2 + Λ 3 + .
analogous to the inequality in (Lemma 13, [37]). Using Equations (35) and (37) we get
Γ 0 ( X ) Γ ( X ) = μ X < p 1 , p 2 , p 3 X υ ( λ 1 p 1 + λ 2 p 2 + λ 3 p 3 + η ) × ( Λ 1 Λ 2 + Λ 3 + + Λ 1 + Λ 2 Λ 3 + + Λ 1 + Λ 2 + Λ 3 2 Λ 1 + Λ 2 + Λ 3 + ) log p 1 log p 2 log p 3 .
Substituting the function υ ( λ 1 p 1 + λ 2 p 2 + λ 3 p 3 + η ) from Equation (38) with its inverse Fourier transform in Equation (32), we get
Γ ( X ) = Υ ( t ) μ X < p 1 , p 2 , p 3 X e ( λ 1 p 1 + λ 2 p 2 + λ 3 p 3 + η ) t log p 1 log p 2 log p 3 × ( Λ 1 Λ 2 + Λ 3 + + Λ 1 + Λ 2 Λ 3 + + Λ 1 + Λ 2 + Λ 3 2 Λ 1 + Λ 2 + Λ 3 + ) d t .
Thus,
Γ ( X ) = Γ 1 ( X ) + Γ 2 ( X ) + Γ 3 ( X ) 2 Γ 4 ( X )
where Γ 1 ( X ) , Γ 2 ( X ) , Γ 3 ( X ) , and Γ 4 ( X ) are the contributions of the consecutive terms on the right side of Equation (39). It is clear that
Γ 1 ( X ) = Γ 2 ( X ) = Υ ( t ) λ 0 X < p 1 , p 2 , p 3 X e ( λ 1 p 1 + λ 2 p 2 + λ 3 p 3 2 + η ) t × Λ 1 Λ 2 + Λ 3 + log p 1 log p 2 log p 3 d t , Γ 3 ( X ) = Υ ( t ) λ 0 X < p 1 , p 2 , p 3 X e ( λ 1 p 1 + λ 2 p 2 + λ 3 p 3 2 + η ) t × Λ 1 + Λ 2 + Λ 3 log p 1 log p 2 log p 3 d t , Γ 4 ( X ) = Υ ( t ) λ 0 X < p 1 , p 2 , p 3 X e ( λ 1 p 1 + λ 2 p 2 + λ 3 p 3 2 + η ) t × Λ 1 + Λ 2 + Λ 3 + log p 1 log p 2 log p 3 d t .
Therefore,
Γ ( X ) 2 Γ 1 ( X ) + Γ 3 ( X ) 2 Γ 4 ( X ) .
We are going to estimate Γ 1 ( X ) . The integrals Γ 3 ( X ) and Γ 4 ( X ) can be treated similarly. Changing the order of summation and bearing in mind Equation (36), we obtain
Γ 1 ( X ) = Υ ( t ) e ( η t ) L ( λ 1 t ) L + ( λ 2 t ) L ˜ + ( λ 3 t ) d t ,
where
L ± ( t ) = d | P ( z ) λ ± ( d ) λ 0 X < p X p + 2 0 ( d ) e ( p t ) log p
and
L ˜ ± ( t ) = d | P ( z ˜ ) λ 3 ± ( d ) λ 0 X < p 2 X p + 2 0 ( d ) e ( p 2 t ) log p ,
We note that λ ± ( d ) are real numbers such that | λ ± ( d ) | 1 . Furthermore, λ ± ( d ) = 0 if 2 | d | or μ ( d ) = 0 . Without utilizing the arithmetic structure of the Rosser weights, we will write them as λ ( d ) . In these cases, instead of L ± ( t ) and L ˜ ± ( t ) , we will simply write L ( t ) and L ˜ ( t ) .
Using the Davenport–Heilbronn adaptation of the circle method (see (Ch. 11, [38])), we divide Γ 1 ( X ) into three separate integrals:
Γ 1 ( X ) = Γ 1 ( 1 ) ( X ) + Γ 1 ( 2 ) ( X ) + Γ 1 ( 3 ) ( X ) ,
where
Γ 1 ( 1 ) ( X ) = | t | Δ Υ ( t ) e ( η t ) L ( λ 1 t ) L + ( λ 2 t ) L ˜ + ( λ 3 t ) d t
Γ 1 ( 2 ) ( X ) = Δ < | t | < H Υ ( t ) e ( η t ) L ( λ 1 t ) L + ( λ 2 t ) L ˜ + ( λ 3 t ) d t
Γ 1 ( 3 ) ( X ) = | t | H Υ ( t ) e ( η t ) L ( λ 1 t ) L + ( λ 2 t ) L ˜ + ( λ 3 t ) d t .
Here, the functions Δ = Δ ( X ) and H = H ( X ) are defined using the expression in Equations (24) and (26). From Lemma 2 [22] we have that
Γ 1 ( 3 ) ( X ) 1
and it is easy to see from Equation (41) that
Γ ( X ) | 2 Γ 1 ( 1 ) ( X ) + Γ 3 ( 1 ) ( X ) 2 Γ 4 ( 1 ) ( X ) | + O | Γ 1 ( 2 ) ( X ) | + O | Γ 3 ( 2 ) ( X ) | + O | Γ 4 ( 2 ) ( X ) | + O 1 .
We will estimate Γ 1 ( 1 ) ( X ) and Γ 1 ( 2 ) ( X ) in Section 4 and Section 5, respectively. The estimation of the sums Γ 3 ( 3 ) ( X ) , Γ 4 ( 3 ) ( X ) , Γ 3 ( 1 ) ( X ) , Γ 4 ( 1 ) ( X ) , Γ 3 ( 2 ) ( X ) , Γ 4 ( 2 ) ( X ) is carried out in the same way. In Section 6, we will finalize the proof of the theorem.

5. Asymptotic Formula for Γ 1 ( 1 ) ( X )

To evaluate the sum Γ 1 ( 1 ) ( X ) , we need asymptotic formulas for L ± ( t ) and L ˜ ± ( t ) when | t | < Δ . Since we will not be using the arithmetic properties of the Rosser weights, we will simply denote them by L ( t ) and L ˜ ( t ) .
In addition, we require the following two estimates.
Lemma 10. 
Let Y 2 , T 2 , D ( log Y ) A for arbitrarily large fixed positive real number A,
L ( D , T , Y ) = d D χ * m o d d | γ | T Y σ ,
and the summation in the inner sum is taken over the non-trivial zeros ρ = σ + i γ of Dirichlet us L-function L ( s , χ ) such that | γ | T . If
D 2 T Y 12 / 25
then
L ( D , T , Y ) Y 4 / 5 D T 1 / 2 ( log Y ) 10 + Y ( log Y ) 14 .
Proof. 
Using Lemmas 6 and 7 and following the same steps as in the proof of Lemma 9 from [39] we get our statement. □
Lemma 11. 
Using notations of Lemma 10 when D 2 T Y 12 / 25 and D ( log Y ) A for arbitrarily large fixed positive real number A and every large enough real Y, the following inequality
L ( D , T , Y ) Y 4 / 5 D T 1 / 2 ( log Y ) 10 + Y ( log Y ) A
is fulfilled.
Proof. 
Using Lemmas 6 and 7 and following the proof of Lemma 9 from [39] we get
L ( T , σ , Y ) Y 1 / 2 T log 2 Y + I 1 log Y + I 2 log Y ,
where
I 1 = 1 / 2 4 / 5 Y t ( D 2 T ) 3 ( 1 t ) 2 t log 10 ( D T ) d t , I 2 = 4 / 5 1 Y t ( D 2 T ) 2 ( 1 t ) 2 log 14 ( D T ) d t .
Using the same reasoning as in the estimation of integral I 1 from (Lemma 9, [39]), we obtain the following
I 1 Y 4 / 5 D T 1 / 2 ( log Y ) 10 .
To estimate the integral I 2 , we note from Lemma 4 that the Dirichlet function L ( s , χ ) does not have zeros in the region.
R e s > 1 C ( ρ ) ( log Y ) A ε .
Choosing ρ = 1 2 A , we find that L ( s , χ ) 0 when Re s > 1 C ( ρ ) log Y . So using Lemma 7 and working as in the estimate of integral I 1 we get
I 2 = 4 / 5 1 C ( ρ ) log Y e h 2 ( t ) d t where h 2 ( t ) = t log Y + ( 2 + ε ) ( 1 t ) log ( D 2 T ) .
As h 2 ( t ) 0 when D 2 T < Y 12 / 25 we obtain
h 2 ( t ) h 2 1 C ( ρ ) log Y = log Y C ( ρ ) log Y 1 12 ( 2 + ε ) 25 log Y C 1 ( ρ , ε ) log Y ,
with C 1 ( ρ , ε ) > 0 . Therefore, for sufficiently large Y
I 2 Y ( log Y ) A .
From Equations (54)–(57) follows the statement of Lemma 11. □
The following Lemma offers information regarding the density of zeros of Riemann’s zeta function.
Lemma 12. 
Let Y 2 , T 2 ,
L ( T , σ , Y ) = | γ | T Y σ
and the summation in the inner sum is taken over the non-trivial zeros ρ = σ + i γ of Riemann’s ζ-function such that | γ | T . Then for enough large Y and T with T < Y 3 / 8 the inequality
L ( T , σ , Y ) Y ( log Y ) A
is fulfilled for an arbitrarily large positive A.
Proof. 
Using Lemmas 5 and 8 and following the proof of Lemma 9 from [39] we get
L ( T , σ , Y ) Y 1 / 2 T log 2 Y + I 1 log 5 Y + I 2 log 5 Y ,
where
I 1 = 1 / 2 3 / 4 Y t T 3 ( 1 t ) 2 t d t , I 2 = 3 / 4 1 c ( log T ) 2 / 3 ( log log T ) 1 / 3 Y t T 8 ( 1 t ) 3 d t .
Working in the same way as in the estimation of integral I 1 in the proof of Lemma 11 when T < Y 25 / 48 we obtain
I 1 Y 3 / 4 T 3 / 5 .
Let us now consider the second integral:
I 2 = 3 / 4 1 c ( log T ) 2 / 3 ( log log T ) 1 / 3 e h 2 ( t ) d t , where h 2 ( t ) = t log Y + 8 ( 1 t ) 3 log T .
It is easy to see that h 2 ( t ) 0 when T < Y 3 / 8 . So,
h 2 ( t ) h 2 1 c ( log T ) 2 / 3 ( log log T ) 1 / 3 = log Y 8 c ( log T ) 1 / 3 3 ( log log T ) 1 / 3 .
and for sufficiently large Y and T
I 2 Y ( log Y ) A .
From Equations (59)–(61) follows the statement of Lemma 12. □
From now on, we will assume that A is a large positive fixed number for which the estimates of Lemmas 10–12 are satisfied.
We shall prove the following.
Lemma 13. 
Let D, D ˜ , Δ, L ( β ) and L ˜ ( β ) , I ( β ) and I ˜ ( β ) , be defined by Equations (24), (25), (28), (29), (42) and (43). If | β | < Δ and Equation (23) are fulfilled, then for large enough X, the following equalities
L ( β ) I ( β ) d D λ ( d ) φ ( d ) = O X ( log X ) A / 2 3 + O X 4 / 5 + ξ / 2 ( log X ) 14 , L ˜ ( β ) I ˜ ( β ) d D ˜ λ ( d ) φ ( d ) = O X 1 / 2 ( log X ) A / 2 3 + O X 2 / 5 + ξ / 2 ( log X ) 14
are fulfilled.
Proof. 
The proof is the same as the proof of Lemma 10 [40] but we use Lemmas 10–12 and the following choice for T and T ˜ :
T = | β | X D ( log x ) A / 2 ; T ˜ = | β | X D ˜ ( log x ) A / 2 .
The following Lemmas provide estimates for the integrals I ( β ) and I ˜ ( β ) , as well as for the integrals derived from them.
Lemma 14. 
For integrals I ( β ) and I ˜ ( β ) , defined by Equations (28) and (29), we have
I ( β ) 1 | β | , I ˜ ( β ) 1 | β | X .
Proof. 
The statement is followed by partial integration. □
Lemma 15. 
Let Δ be defined by Equation (24). Then for integrals I ( β ) and I ˜ ( β ) , defined by Equations (28) and (29), we have
Δ Δ | I ( β ) | 2 d β X and Δ Δ | I ˜ ( β ) | 2 d β 1 .
Proof. 
The proof of the first inequality is similar to that of Lemma 11 [39]. We will demonstrate the second inequality. We notice that
Δ Δ | I ˜ ( β ) | 2 d β 0 Δ | I ˜ ( β ) | 2 d β = 0 ϑ | I ˜ ( β ) | 2 d β + ϑ Δ | I ˜ ( β ) | 2 d β .
Using Equation (63) and the basic estimate | I ˜ ( β ) | X , we can conclude that
Δ Δ | I ˜ ( β ) | 2 d β ϑ X + ϑ Δ 1 X β 2 d β ϑ X + 1 ϑ X .
By choosing ϑ = 1 X , we prove the second inequality in our statement. □
The following Lemma is analogous to Lemma 11 from [39].
Lemma 16. 
Let Δ, D, D ˜ be defined by Equations (24), (25) and (23) are fulfilled. Then for sums L ( β ) and L ˜ ( β ) (see Equations (42) and (43)), we have
Δ Δ | L ( β ) | 2 d β X log 5 X and Δ Δ | L ˜ ( β ) | 2 d β X log 5 X .
Proof. 
Using the inequality
Δ Δ | L ( β ) | 2 d β 1 / 2 1 / 2 | L ( β ) | 2 d β ,
the definition of Equation (42) and arguing as in §6 [22] we obtain
Δ Δ | L ( β ) | 2 d β X log 5 X .
Working in a similar way, we get the estimate for the second integral. □
From now on, we will put
B ± = B ± ( D , z ) , B ˜ ± = B ± ( D ˜ , z ˜ ) .
To find asymptotic formulae for Γ 1 ( 1 ) ( X ) we need the following
Lemma 17. 
Let Δ, D, D ˜ , z, z ˜ , I ( β ) , I ˜ ( β ) , L ( β ) and L ˜ ( β ) be defined by Equations (24), (25), (27)–(29), (42), (43) and
R = Δ Δ Y ( t ) e ( η t ) B ± B ± B ˜ ± I ( λ 1 t ) I ( λ 2 t ) I ˜ ( λ 3 t ) L ± ( λ 1 t ) L ± ( λ 2 t ) L ˜ ± ( λ 3 t ) d t .
Then
R E X 3 / 2 ( log X ) A / 2 3 + X 14 + 5 ξ 10 ( log X ) 14 .
Proof. 
From Equation (33) follows
R E J 1 + J 2 + J 3 ,
where
J 1 = 0 Δ | B ± I ( λ 1 t ) L ± ( λ 1 t ) | | B ± I ( λ 2 t ) | | B ˜ ± I ˜ ( λ 3 t ) | d t , J 2 = 0 Δ | L ± ( λ 1 t ) | | B ± I ( λ 2 t ) L ± ( λ 2 t ) | | B ˜ ± I ˜ ( λ 3 t ) | d t , J 3 = 0 Δ | L ± ( λ 1 t ) | | L ± ( λ 2 t ) | | B ˜ ± I ˜ ( λ 3 t ) L ˜ ± ( λ 3 t ) | d t .
From Lemma 13, and Equations (19), (20) and (23) we have
J 1 X ( log X ) A / 2 2 + X 4 / 5 + ξ / 2 ( log X ) 13 0 Δ | I ( λ 2 t ) | | I ˜ ( λ 3 t ) | d t .
Using the Cauchy-Schwarz inequality and Lemma 15, we get
J 1 X ( log X ) A / 2 3 + X 4 / 5 + ξ / 2 ( log X ) 14 × 0 Δ | I ( λ 2 t ) | 2 d t 1 / 2 0 Δ | I ˜ ( λ 3 t ) | 2 d t 1 / 2 X 3 / 2 ( log X ) A / 2 3 + X 13 + 5 ξ 10 ( log Y ) 14 .
Using analogous reasoning, Lemma 16 and Equation (23) we get
J 2 X 3 / 2 ( log X ) A / 2 3 + X 13 + 5 ξ 10 ( log X ) 14 , J 3 X 3 / 2 ( log X ) A / 2 3 + X 14 + 5 ξ 10 ( log X ) 14 .
From Equations (65)–(67) follows the statement of the Lemma. □
From Lemma 17 and from Equation (45) we get
Γ 1 ( 1 ) ( X ) = B B + B ˜ + | t | Δ Υ ( t ) e ( η t ) I ( λ 1 t ) I ( λ 2 t ) I ˜ ( λ 3 t ) d t + O E X 3 / 2 ( log X ) A / 2 3 + O E X 14 + 5 ξ 10 ( log X ) 14
Let
J ( X ) = Υ ( t ) e ( η t ) I ( λ 1 t ) I ( λ 2 t ) I ˜ ( λ 3 t ) d t , J 1 ( X ) = Δ Υ ( t ) e ( η t ) I ( λ 1 t ) I ( λ 2 t ) I ˜ ( λ 3 t ) d t , J 2 ( X ) = Δ Υ ( t ) e ( η t ) I ( λ 1 t ) I ( λ 2 t ) I ˜ ( λ 3 t ) d t .
We will evaluate J 1 ( X ) . The estimate of the integral J 2 ( X ) is the same. Using Lemma 14 and Equation (33) we get
J 1 ( X ) E X Δ d t | λ 1 λ 2 λ 3 | t 3 E Δ 2 X .
Using the restriction in Equation (23), we choose
ξ = 1 100 , D = 47 300 ε and D ˜ = 23 300 ε ,
where ε is an arbitrarily small positive number. From Equations (19), (20), (68) and (69) we get
Γ 1 ( 1 ) ( X ) = J ( X ) B B + B ˜ + + O E X 3 / 2 ( log X ) A / 2 3 .
Reasoning as in Lemma 4 [22] for
λ 0 < min λ 1 4 | λ 3 | , λ 2 4 | λ 3 | , 1 16 ,
we have
J E X 3 / 2 ,
with a constant implied by the symbol ≫ depending only on λ 1 , λ 2 and λ 3 .

6. Asymptotic Formula for Γ 1 ( 2 ) ( X )

From Equation (49), we see that to find a nontrivial lower bound for Γ we have to prove that the integrals Γ 1 ( 2 ) , Γ 3 ( 2 ) and Γ 4 ( 2 ) are small enough. To establish this, we will use the fact that the ratio λ 1 / λ 2 is an irrational number. This will allow us to show that one of the sums L ± ( λ 1 t ) or L ± ( λ 2 t ) can always be estimated non-trivially. By the restrictions in Equation (23), it follows that D X 4 / 25 and for these D we will use Lemma 9. From this Lemma, we see that if
q X 1 / 3 , X 2 / 3
and if λ i t is irrational, then
L ± ( λ i t ) max λ 0 X Y X | W ( Y ) | X 5 / 6 + ε .
Let
V ( t , X ) = min L ± ( λ 1 t ) , L ± ( λ 2 t ) .
Also, we need the following
Lemma 18. 
Let t , X , λ 1 , λ 2 R ,
| t | ( Δ , H ) ,
where Δ and H are denoted by Equations (24) and (26), λ 1 , λ 2 satisfy Equation (3) and V ( t , X ) is defined by Equation (74). Then there exists a sequence of real numbers X 1 , X 2 , such that
V ( t , X j ) X j 5 / 6 + ε , j = 1 , 2 , .
Proof. 
The proof is the same as in Lemma 8, [24]. Since λ 1 λ 2 R / Q by (Corollary 1B, [41]), there exist infinitely many fractions a q with arbitrarily large denominators such that
| λ 1 λ 2 a q | < 1 q 2 , ( a , q ) = 1 .
For sufficiently large q, we choose X such that
q = X 2 / 3 .
Following the proof of Lemma 8, [24], we get an infinite sequence q ( 1 ) , q ( 2 ) , of values of q, satisfying Equation (77). Then using Equation (78) one gets an infinite sequence X 1 , X 2 , of values of X, such that at least one of the numbers λ 1 t and λ 2 t can be approximated by rational numbers with denominators, satisfying Equation (72). Hence, the inequality Equation (73) is fulfilled, and the proof is completed. □
The following Lemma gives an upper bound for the number of integers that can be represented as the sum of two squares belonging to some arithmetical progressions in two different ways.
Lemma 19. 
Let d i D , d i N , a i Z , i = 1 , . . . , 4 , D 2 , 0 < λ 0 < 1 and
T ( X ) = d i D i = 1 , 2 , 3 , 4 λ 0 X < n i 2 X n 1 2 + n 2 2 = n 3 2 + n 4 2 λ 0 X < n 1 2 + n 2 2 X n i a i ( d i ) i = 1 , 2 , 3 , 4 1 .
Then T ( X ) X log X .
Proof. 
It is well known that the number of representations of the integer n as a sum of two squares is τ ( n ) . Using this fact, we obtain
T ( X ) d i D | a i | d i i = 1 , 2 n i X n 1 2 + n 2 2 X n i a i ( d i ) i = 1 , 2 τ ( n 1 2 + n 2 2 ) d i X | a i | d i i = 1 , 2 n X n = n 1 2 + n 2 2 n i a i ( d i ) i = 1 , 2 τ ( n ) n X n = n 1 2 + n 2 2 τ ( n ) n X τ ( n ) .
It is well known (see Equation [42]) that n X τ ( n ) X log X and from here the Lemma assertion follows. □
To estimate the integral Γ 1 ( 2 ) we will use Equation (74) to notice that
| L ( λ 1 t ) | | L + ( λ 2 t ) | | V ( t , X j ) | 1 2 | L ( λ 1 t ) | 1 2 | L + ( λ 2 t ) | , | L ( λ 1 t ) | | L + ( λ 2 t ) | | V ( t , X j ) | 1 2 | L ( λ 1 t ) | | L + ( λ 2 t ) | 1 2 .
Next, from Equation (33), above inequalities and estimate Equation (76) for integral Γ 1 ( 2 ) ( X j ) , denoted by Equation (46) we find
Γ 1 ( 2 ) ( X j ) E | V ( t , X j ) | 1 2 I 1 ( X j ) + I 2 ( X j ) E X j 5 / 12 + ε / 2 I 1 ( X j ) + I 2 ( X j ) ,
where
I 1 ( X j ) = Δ H | L ( λ 1 t ) | 1 2 | L + ( λ 2 t ) | | L ˜ + ( λ 3 t ) | d t , I 2 ( X j ) = Δ H | L ( λ 1 t ) | | L + ( λ 2 t ) | 1 2 | L ˜ + ( λ 3 t ) | d t .
We will estimate only the integral I 1 ( X j ) . The estimation of I 2 ( X j ) is the same. Using twice the Cauchy-Schwarz inequality, we get
I 1 ( X j ) Δ H L + ( λ 2 t ) 2 d t 1 2 Δ H | L ( λ 1 t ) | 2 d t 1 4 Δ H L ˜ + ( λ 3 t ) 4 d t 1 4 .
Arguing as in §6 [22] we obtain
Δ H L ± ( λ j t ) 2 d t H X j log 5 X j .
From Equation (43) follows
Δ H L ˜ + ( λ 3 t ) 4 d t 0 [ λ 3 ] H + 1 L ˜ + ( λ 3 t ) 4 d t H d i | P ( z ) i = 1 , 2 , 3 , 4 ( λ 0 X j ) 1 2 < p j X j 1 2 p j + 2 0 ( d j ) j = 1 , 2 , 3 , 4 log p 1 log p 2 log p 3 log p 4 0 1 e ( ( p 1 2 + p 2 2 p 3 2 p 4 2 ) y ) d y H log 4 X j d i | P ( z ) i = 1 , 2 , 3 , 4 λ 0 X j < p j X j p 1 2 + p 2 2 = p 3 2 + p 4 2 p j + 2 0 ( d j ) j = 1 , 2 , 3 , 4 1 H T ( X j ) log 4 X j ,
with T ( X j ) defined by Equation (79). From Lemma 19 we have T ( X j ) X j log X j . Therefore,
Δ H L ˜ + ( λ 3 t ) 4 d t H X j log 5 X j .
From Equations (26) and (80)–(83) with θ > ε / 2 follows
Γ 1 ( 2 ) ( X j ) E H X j 17 / 12 + ε / 2 log 5 X j E X j 3 / 2 ( log X j ) A / 2 3 .
Now from Equations (44), (48), (71) and (84) we obtain
Γ 1 ( X j ) = J ( X j ) B B + B ˜ + + O E X j 3 / 2 ( log X j ) A / 2 3 .
Similarly, we can determine Γ 3 ( X j ) and Γ 4 ( X j ) . From Equations (41) and (85) we get
Γ ( X j ) J ( X j ) 2 B B + B ˜ + + B + 2 B ˜ 2 B + 2 B ˜ + .
Using Equations (16), (17) and (19) we obtain
Γ ( X j ) J ( X j ) B 2 ( z ) B ( z ˜ ) 2 f ( s ) + f ( s ˜ ) 2 F 2 ( s ) F ( s ˜ ) ,
where f and F are functions of the linear sieve as described in Lemma 1. Choosing s = s ˜ = 3.2825 and Equation (18) we get
Γ ( X j ) 0.000178 J ( X j ) B 2 ( z ) B ( z ˜ ) .
From Equation (70) with ε = 0.00001 and
s = log D log z , s ˜ = log D ˜ log z ˜ .
we receive 1 α = 20.95 and 1 α ˜ = 42.81 . Therefore,
p 1 + 2 = P 20 , p 2 + 2 = P 20 , p 3 + 2 = P 42
and the proof of Theorem 1 is complete.

7. Conclusions

Diophantine inequalities are an essential topic in analytic number theory. In general, we consider the following problem. Let r be a positive integer and let λ 1 , , λ r be non-zero real numbers, while k 1 , , k r be positive real numbers. Furthermore, let η be a real number. The goal is to demonstrate that the inequality
| λ 1 p 1 k 1 + λ 2 p 2 k 2 + + λ r p r k r + η | < E
has infinitely many solutions with primes p 1 , , p r , where E can be made as small as possible. The number of variables r is significant, and there are some hypotheses about the irrationality of at least one ratio λ i / λ j . Furthermore, the assumption that the numbers λ i do not share the same sign is crucial. When we impose some restrictions on the prime numbers p 1 , , p r in Equation (87), we can obtain a wide variety of problems regarding the solvability of Diophantine inequalities. If, in addition to the restrictions on p 1 , , p r , we also add restrictions on the number r and the variety of exponents k i , we get a large number of interesting problems related to Diophantine inequalities. As a rule, the approach to each of these problems combines the Davenport–Heilbronn adaptation of the circle method with a method that corresponds to the specific restrictions imposed on the participating prime numbers.

Author Contributions

Conceptualization, A.G. and T.L.T.; methodology, A.G. and T.L.T.; validation, A.G. and T.L.T.; formal analysis, A.G. and T.L.T.; writing—original draft preparation, A.G. and T.L.T.; writing—review and editing A.G. and T.L.T.; funding acquisition A.G. and T.L.T. All authors have read and agreed to the published version of the manuscript.

Funding

This study is financed by the European Union-Next Generation EU through the National Recovery and Resilience Plan of the Republic of Bulgaria, project DUECOS BG-RRP-2.004-0001-C01.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

The original contributions presented in this study are included in the article. Further inquiries can be directed to the corresponding author.

Conflicts of Interest

The authors declare no conflicts of interest.

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Todorova, T.L.; Georgieva, A. A Diophantine Inequality Involving Mixed Powers of Primes with a Specific Type. Mathematics 2025, 13, 3065. https://doi.org/10.3390/math13193065

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Todorova TL, Georgieva A. A Diophantine Inequality Involving Mixed Powers of Primes with a Specific Type. Mathematics. 2025; 13(19):3065. https://doi.org/10.3390/math13193065

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Todorova, Tatiana L., and Atanaska Georgieva. 2025. "A Diophantine Inequality Involving Mixed Powers of Primes with a Specific Type" Mathematics 13, no. 19: 3065. https://doi.org/10.3390/math13193065

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Todorova, T. L., & Georgieva, A. (2025). A Diophantine Inequality Involving Mixed Powers of Primes with a Specific Type. Mathematics, 13(19), 3065. https://doi.org/10.3390/math13193065

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