Diophantine approximation with one prime, two squares of primes and one $k$-th power of a prime

Let $1<k<14/5$, $\lambda_1,\lambda_2,\lambda_3$ and $\lambda_4$ be non-zero real numbers, not all of the same sign such that $\lambda_1/\lambda_2$ is irrational and let $\omega$ be a real number. We prove that the inequality $|\lambda_1p_1+\lambda_2p_2^2+\lambda_3p_3^2+\lambda_4p_4^k-\omega|\le (\max_j p_j)^{-\frac{14-5k}{28k}+\varepsilon}$ has infinitely many solutions in prime variables $p_1,p_2,p_3,p_4$ for any $\varepsilon>0$.

has infinitely many solutions in prime variables p 1 , p 2 , p 3 , p 4 for any ε > 0.
Many recent such results are known with various types of assumptions and conclusions.Many of them deal with the number of exceptional real numbers ω such that the inequality r − ω| ≤ η has no solution in prime variables p 1 , . . ., p r , for small η > 0 fixed.
Brüdern, Cook and Perelli in [BCP] dealt with binary linear forms in prime arguments, Cook and Fox in [CF] dealt with a ternary form with squares of primes that was improved in term of approximation by Harman in [H ].Cook in [C] gave a more general description of the problem, later improved by Cook and Harman in [CH].
There are some differences between the results quoted above and our purpose: in our case the value of η does depend on the primes p j and it will be actually a negative power of the maximum of the p j while in the papers quoted above η is a small negative power of ω.In their papers the assumption that the coefficients λ j are all positive is not a restriction.Moreover k j is the same positive integer for all j.Nevertheless the assumption that λ 1 /λ 2 must be irrational is still the heart of the matter.
Vaughan in [V ] follows another approach, that is the same we are using in our article: dealing with a ternary linear form in prime arguments and assuming some more suitable conditions on the λ j , he proved that there are infinitely many solutions of the problem when η depends on the maximum of the p j ; in his case η = (max j p j ) − 1 10 .Such result was improved by Baker and Harman in [BH] with exponent − 1 6 , by Harman in [H ] with exponent − 1 5 and finally by Matomäki in [Ma] with exponent − 2 9 .Languasco and Zaccagnini in [LZ ] and [LZ ] dealt with a ternary problem with a k-th power of a prime.In this case the value of η is a negative power of the maximum of the p j also depending on the parameter k: the idea in this case is to get both the widest k-range and the strongest bound for the approximation.
Languasco and Zaccagnini dealt also with a quaternary form in [LZ ] with a prime and squares of primes obtaining η = (max j p j ) − 1 18 ; this was improved by Liu & Sun in [LSu] with η = (max j p j ) − 1 16 using the Harman technique.Wang & Yao in [WY] improved the approximation to the exponent − 1 14 ; in this paper we generalized the problem to a real power k ∈ 1, 14 5 .
. O We use a variant of the classical circle method that was introduced by Davenport and Heilbronn in [DH] in order to attack this kind of Diophantine problems.The integration on a circle, or equivalently on the interval [0, 1], is replaced by integration on the whole real line.
Throughout this paper p i denotes a prime number, k ≥ 1 is a real number, ε is an arbitrarily small positive number whose value could vary depending on the occurrences and ω is a fixed real number.In order to prove that ( ) has infinitely many solutions, it is sufficient to construct an increasing sequence X n that tends to infinity such that ( ) has at least one solution with max p j ∈ [δX n , X n ], with δ > 0 fixed, depending on the choice of λ j .Let q be a denominator of a convergent to λ 1 /λ 2 and let X n = X (dropping the suffix n) run through the sequence X = q 7/3 .Set where e(α) = e 2πiα .We will approximate S k with T k and U k .
By the Prime Number Theorem and first derivative estimates for trigonometric integrals we have Moreover the Euler summation formula implies that We also need a continuous function we will use to detect the solutions of ( ), so we introduce whose inverse Fourier transform is for α 0 and, by continuity, K η (0) = η 2 .It vanishes at infinity like |α| −2 and in fact it is trivial to prove that The original works of Davenport-Heillbronn in [DH] and later Vaughan in [V ] and [V ] approximate directly the difference |S k (α) − T k (α)|, estimating it with O(1) using the Euler summation formula.The L 2 -norm estimation approach (see [BCP] and [LZ ]) improves these estimation taking the L 2 -norm of |S k (α) − T k (α)| leading to significantly better conditions and to have a wider major arc compared to the original DH approach.In fact, setting the generalized version of the Selberg integral we have the following lemmas.

Lemma ([LZ
Lemma ([LZ ], Theorem ).Let k ≥ 1 be a real number and ε be an arbitrarily small positive constant.There exists a positive constant c 1 (ε), which does not depend on k, such that and let us define where X is a measurable subset of R.
From the definitions of the S j (λ i α) and performing the Fourier transform for K η (α), we get where N(X) actually denotes the number of solutions of the inequality ( ) with (p 1 , p 2 , p 3 , p 4 ) ∈ P(X).In other words I(η, ω, R) provides a lower bound for the quantity we are interested in; therefore it is sufficient to prove that I(η, ω, R) > 0.
We now decompose R into subsets such that R = M ∪ m ∪ t where M is the major arc, m is the minor arc (or intermediate arc) and t is the trivial arc.The decomposition is the following: The parameters P = P(X) > 1 and R = R(X) > 1/η are chosen later (see ( ) and ( )) as well as η = η(X), that, as we explained before, we would like to be a small negative power of max p j (and so of X, see ( )).
We are expecting to have on M the main term with the right order of magnitude without any special hypothesis on the coefficients λ j .It is necessary to prove that I(η, ω, m) and I(η, ω, t) are both o(I(η, ω, M)): the contribution from the trivial is "tiny" with respect to the main term.The real problem is on the minor arc where we will need the full force of the hypothesis on the λ j and the theory of continued fractions.
Remark: from now on, anytime we use the symbol ≪ or ≫ we drop the dependence of the approximation from the constants λ j , δ and k.
. .Lemmas.In this paper we will also use Lemmas --of [GLZ] that allow us to have an estimation of mean value of |S k (α)| 4 : Lemma ([GLZ], Lemma ).Let ε > 0 fixed, k > 1, γ > 0 and let A(X 1/k ; k; γ) denote the number of solutions of the inequality Finally, we will use the following Lemma.

Lemma . ∫
Proof.The first statement comes directly from Prime Number Theorem, while the second estimation is based on Satz of [R], p. .

. T
Let us start from the major arc and the computation of the main term.We replace all S k defined in ( ) with the corresponding T k defined in ( ).This replacement brings up some errors that we must estimate by means of Lemma , the Cauchy-Schwarz and the Hölder inequalities.We write say.For brevity, since the computations for J 2 and J 3 are similar to, but simpler than, the corresponding ones for J 4 and J 5 , we will leave them to the reader.
. .Main Term: lower bound for J 1 .As the reader might expect the main term is given by the summand J 1 . Let Using inequalities ( ) and ( ), Apart from trivial changes of sign, there are essentially three cases as in [LZ ]: We deal with the second case, the other ones being similar: let us perform the following change of variables: 4 , so that the set D becomes essentially [δX, X] 4 .Let us define D ′ = [δX, (1 − δ)X] 4 for large X, as a subset of D. The Jacobian determinant of the change of variables above is so that, for every choice of (u 1 , u 2 , u 3 ) the interval Finally, , which is the expected lower bound.
. .Bound for J 4 .The computations on J 2 and J 3 are similar to and simpler than the corresponding one on J 4 , so we will skip it.Using the triangle inequality, say, where U 2 (λ 3 α) is given by ( ).
Using the trivial inequalities |S 2 (α)| ≪ X 1 2 , ( ) and then the Cauchy-Schwarz inequality, for any fixed A (Lemmas and ), we have as long as A > 1. Again using ( ) and ( ), Remembering that |α| ≤ P X on M and using the Hölder inequality, trivial bounds and Lemma , we have Since we must have is sufficient for our purpose.
. .Bound for J 5 .In order to provide an estimation for J 5 , we use ( ), and then the arithmetic-geometric inequality (ab ≤ a 2 + b 2 ): The two terms are equivalent; then we consider only one of them say.Using trivial estimates, then using the Hölder inequality, for any fixed A > 2 by Lemmas and we have Now we turn to B 5 : by ( ) we have Using trivial estimates and Lemma Then we need Collecting all the bounds for P, that is, ( ), ( ), ( ) we can take In fact, if we consider ( ) and ( ), we should choose the most restrictive condition between the two: if k ≤ 25 12 , P = X .T By the arithmetic-geometric mean inequality and the trivial bound for S k (λ 4 α), we see that say.Using the PNT and the periodicity of S 1 (α), we have Collecting ( ) and ( ), Hence, remembering that |I(η, ω, t)| must be o η 2 X 1 k +1 , i.e. of the main term, the choice . T In [LZ ] Lemma it is proven that the measure of the set where |S 1 (λ 1 α)| 1 2 and |S 2 (λ 2 α)| are both large for α ∈ m is small, exploiting the fact that the ratio λ 1 /λ 2 is irrational.
Lemma (Vaughan [V ], Theorem .).Let α be a real number and a, q be positive integers satisfying (a, q) = 1 and α − a q < 1 q 2 .Then We now state some considerations about Lemmas : Corollary (Liu-Sun [LSu], Corollary .).Suppose that X ≥ Z ≥ X 1− 1 5 +ε and |S 1 (λ 1 α)| > Z. Then there are coprime integers (a, q) = 1 satisfying Then there are coprime integers (a, q) = 1 satisfying Let us now split m into two subsets m and m * = m\ m.In turn m = m 1 ∪ m 2 , where Using the Hölder inequality, Lemma and the definition of m 1 we obtain Using the Hölder inequality, Lemma and the definition of m 2 we obtain η 2 = R.Following the dyadic dissection argument as in [H ] we divide m * into disjoint sets E(Z 1 , Z 2 , y) in which, for α ∈ E(Z 1 , Z 2 , y), we have where 16 +ε and y = 2 k 3 X − 3 4 −ε for some non-negative integers k 1 , k 2 , k 3 .
It follows that the disjoint sets are, at the most, ≪ log 3 X.Let us define A as a shorthand for the set E(Z 1 , Z 2 , y); we have the following result about the Lebesgue measure of A following the same lines of Lemma in [Mu]: Lemma .We have µ(A) ≪ yX 18 7 +6ε Z −2 1 Z −4 2 , where µ(•) denotes the Lebesgue measure.Proof.If α ∈ A, by Corollaries and Lemma there are coprime integers (a 1 , q 1 ) and (a 2 , q 2 ) such that m 2 |S k (λ 4 α)| 4 K η (α)dα) and ( ) must be o η 2 X 1+ 1 k , consequently it is clear that for 1 < k < 2, η is a negative power of Xindependently from the value of k.The we have the following most restrictive condition for k ≥ 2: η = ∞ X − 15−5k 28k +ε .It remains to discuss the set m * in which the following bounds hold simultaneously |S 1 (λ 1 α)| > X 6 7 +ε , |S 2 (λ 2 α)