Diophantine approximation with one prime, two squares of primes and one kth power of a prime

2.1 Setting the problem

Let

and let us define

where X is a measurable subset of R .

From the construction of ρ ( m ) follows that, if ω ( m ) is the characteristic function of the set of primes,

Then, from the definitions of S j ( λ i α ) and S ˜ 2 ( λ 2 α ) , and performing the Fourier transform for K η ( α ) , we get

where N ( X ) denotes the number of solutions of the inequality (1) 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:

so that ℐ ( η , ω , R ) = ℐ ( η , ω , ℳ ) + ℐ ( η , ω , m ) + ℐ ( η , ω , t ) .

The parameters P = P ( X ) > 1 and R = R ( X ) > 1 / η are chosen later (see (13) and (16)) as well as η = η ( X ) , that, as we explained before, we would like to get a small negative power of max p j (and so of X , see (24)).

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 .

2.2 Lemmas

In this paper, we will also use Lemmas 3-4-10 of [20] and (2.5) of [15] that allow us to have an estimation of mean value of ∣ S k ( α ) ∣ 4 and ∣ S ˜ 2 ( α ) ∣ 4 :

Finally, we will use the following Lemma.

3.1 Main term: lower bound for J 1

As the reader might expect the main term is given by the summand J 1 .

Let H ( α ) = T 1 ( λ 1 α ) T 2 ( λ 2 α ) T 2 ( λ 3 α ) T k ( λ 4 α ) K η ( α ) e ( − ω α ) so that

Using inequalities (9) and (7),

provided that P → + ∞ . Let D = [ δ X , X ] × [ ( δ X ) 1 2 , X 1 2 ] 2 × [ ( δ X ) 1 k , X 1 k ] ; we have

Apart from trivial changes of sign, there are essentially three cases:

1. λ 1 > 0 , λ 2 > 0 , λ 3 > 0 , λ 4 < 0 ;

2. λ 1 > 0 , λ 2 > 0 , λ 3 < 0 , λ 4 < 0 ;

3. λ 1 > 0 , λ 2 < 0 , λ 3 < 0 , λ 4 < 0 .

We deal with the second case, cases 1 and 3 being similar. Let us perform the following change of variables: u 1 = t 1 − ω λ 1 , u 2 = t 2 2 , u 3 = t 3 2 , u 4 = t 4 k , so that the set D becomes essentially [ δ X , X ] 4 . Let us define D ′ = [ δ X , ( 1 − δ ) X ] 4 for large X , as a subset of D . The Jacobian determinant of the change of variables above is 1 4 k u 2 − 1 2 u 3 − 1 2 u 4 1 k − 1 . Then

Now, for j = 1 , 2 , 3 let a j = 4 ∣ λ 4 ∣ ∣ λ j ∣ , b j = 3 2 a j and D j = [ a j δ X , b j δ X ] ; if u j ∈ D j , then

so that, for every choice of ( u 1 , u 2 , u 3 ) the interval

is contained in [ δ X , ( 1 − δ ) X ] . In other words, for u 4 ∈ [ a , b ] the values of λ 1 u 1 + λ 2 u 2 + λ 3 u 3 + λ 4 u 4 cover the whole interval [ − η , η ] . Hence, for any ( u 1 , u 2 , u 3 ) ∈ D 1 × D 2 × D 3 we have

Finally,

which is the expected lower bound.

3.2 Bound for J 3

From partial summation on (6) we get

then

Finally, J 3 can be estimated as follows:

using (7).

3.3 Bound for J 4

Using the triangle inequality,

where U 2 ( λ 3 α ) is given by (3).

Using (7) and the Cauchy-Schwarz inequality,