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Green and Tao famously proved (The primes contain arbitrarily long arithmetic progressions) that there are arbitrarily long arithmetic progressions in the primes. Specifically, for $k = 3$ this implies that there are infinitely many integers $x,y$ such that

$$\Lambda(x-2y) \Lambda(x) \Lambda(x+2y) > 0$$

where $\Lambda$ is the von Mangoldt function. However, as far as I know their result is somewhat weak quantitatively, in the sense that it is far from producing an asymptotic formula for the expression above.

My question is about the asymptotic evaluation of the sum

$$\displaystyle \sum_{\substack{|x(x^2 - 4y^2)| \leq X \\ x,y \in \mathbb{Z}}} \Lambda(x-2y) \Lambda(x) \Lambda(x+2y),$$

as a function of $X$.

If we replace each of the von Mangoldt functions with $1$, then there is an asymptotic formula for this sum, given for example by Mahler's theorem. In particular we have

$$\displaystyle \sum_{\substack{|x(x^2 - 4y^2)| \leq X \\ x,y \in \mathbb{Z}}} 1 = \frac{3\Gamma^2(1/3)}{2^{4/3}\Gamma(2/3)} X^{\frac{2}{3}} + O \left(X^{\frac{1}{2}} \right),$$

see for example my paper with Cam Stewart for a summary (On the representation of integers by binary forms).

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    $\begingroup$ This particular sum can be handled by the classical circle method in the bulk region where $x, x-2y, x+2y \asymp X^{1/3}$, and some standard upper bound sieve to handle the tail region where $x-2y = o(X^{1/3})$. For longer progressions, one can use the asymptotics in my later paper with Ben mathscinet.ams.org/mathscinet/article?mr=2680398 . There are now also estimates that can handle narrow arithmetic progressions, e.g., Theorem 1.7 of arxiv.org/abs/2204.03754 . $\endgroup$
    – Terry Tao
    Commented Aug 10, 2023 at 17:37

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