## The prime $$k$$-tuplets in arithmetic progressions.(English)Zbl 0797.11076

Let $$k \geq 2$$, $$a_ j$$ be non-zero integers and $$b_ j$$ be integers for $$0 \leq j \leq k-1$$. Put $${\mathfrak a} = (a_ 0, \dots, a_{k-1}, b_ 0)$$, $${\mathfrak b} = (b_ 1, \dots, b_{k-1})$$, $N(x,{\mathfrak b}) = \{n:1 \leq a_ j n + b_ j \leq x, \quad j = 0,\dots, k-1\},$
$\Psi (x,{\mathfrak b},a,q) = \sum_{{n \in N (x,{\mathfrak b}) \atop n \equiv a \pmod q}} \prod^{k-1}_{j=0} \wedge (a_ jn + b_ j),$ $$Z(x) = \{b:| N (x,{\mathfrak b}) | \neq 0\}$$ and consider the inequality $\sum_{q \leq Q} \max_{1 \leq a \leq q} \sum_{{\mathfrak b} \in Z(x)} | \Psi (x,{\mathfrak b}, a,q) - \text{ expected main term} | \ll x^ k (\log x)^{-A} \tag{1}$ for fixed $${\mathfrak a}$$ and any fixed $$A>0$$. Using the circle method, A. Balog [Analytic number theory, Prog. Math. 85, 47-75 (1990; Zbl 0719.11066)] proved that (1) holds for any $$h \geq 2$$ when $$Q \leq x^{1/3} (\log x)^{-B}$$, $$B=B(A)>0$$, and H. Mikawa [Tsukuba J. Math. 10, 377-387 (1992; Zbl 0778.11053)] improved Balog’s result to $$Q \leq x^{1/2} (\log x)^{-B}$$, $$B = B(A)>0$$, in the case $$h=2$$.
The author extends Mikawa’s result to the general case $$k \geq 2$$, i.e. he proves (1) for any fixed $${\mathfrak a}$$ and $$A>0$$, $$k \geq 2$$ and $$Q \leq x^{1/2} (\log x)^{-B}$$, $$B = B(A,k)>0$$. He also proves a short intervals version of (1), where $$N(x,{\mathfrak b})$$ is replaced by $$N(x,y,{\mathfrak b}) = \{n:x-y<a_ jn + b_ j\leq x$$, $$j = 0,\dots, k-1\}$$ and analogously for $$\Psi (x,y, {\mathfrak b}, a,q)$$ and $$2(x,y)$$, and the right hand side of (1) is replaced by $$y^ k (\log x)^{-A}$$, provided $$x^{2/3} (\log x)^ c<y \leq x$$ and $$Q \leq yx^{-1/2} (\log x)^{- B}$$, $$B=B(A,k)>0$$.
Reviewer: A.Perelli (Genova)

### MSC:

 11P32 Goldbach-type theorems; other additive questions involving primes 11N13 Primes in congruence classes

### Keywords:

Prime $$k$$-tuplets on average

### Citations:

Zbl 0719.11066; Zbl 0778.11053
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