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On a ternary quadratic form over primes. (English) Zbl 1243.11093

Let \(A(x,y,z) = xy + yz + zx\). It is natural to conjecture that every sufficiently large integer \(N\) with \(N \equiv 3, 11 \pmod {12}\) can be expressed as \(N = A(p_1, p_2, p_3)\), where \(p_1, p_2\) and \(p_3\) are primes. However, while natural, such a conjecture appears to lie far beyond the reach of present methods. In the paper under review, the authors, motivated by the above conjecture, study the multiplicative structure of the sequence \[ \mathcal A(X) = \{ A(p_1, p_2, p_3) : X < p_1, p_2, p_3 \leq 2X \}. \] Let \(\mathcal A_q(X)\) denote the subsequence of \(\mathcal A(X)\) consisting of those elements of \(\mathcal A(X)\) that are divisible by an integer \(q\). The main result of the paper is an asymptotic formula for the number of elements of \(\mathcal A_q(X)\), with a large uniformity in \(q\). In particular, it follows from it that \[ \#\mathcal A_q(X) \sim \prod_{p \mid q} \left( 1 - \frac 1{(p-1)^2} \right) \frac {X^3}{q(\log X)^3} \] for most odd \(q\), with \(Q < q \leq 2Q\) and \(Q \leq X^{17/16-\varepsilon}\). The authors then use various sieve ideas to obtain lower bounds for the number of elements of \(\mathcal A(X)\) that have at most two prime divisors and for the number of elements of \(\mathcal A(X)\) that have a prime divisor \(p > X^{11/10}\).
The authors derive the above arithmetic results from new bounds on the exponential sum \[ S_q(a; x) = \sum_{_{\substack{ x < p \leq 2x\\(p,q)=1 }}} \exp(2\pi ia\bar p/q), \] where the summation is over primes \(p\) and \(\bar p\) denotes the multiplicative inverse of \(p\) modulo \(q\). In particular, they prove that if \((a, q) = 1\) and \(q \geq x^{3/4}\), then \[ S_q(a; x) \ll \big( x^{15/16} + q^{1/4}x^{2/3} \big)q^{\varepsilon} \] for any fixed \(\varepsilon > 0\).

MSC:

11N25 Distribution of integers with specified multiplicative constraints
11L20 Sums over primes
11N35 Sieves
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