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Consider all Pythagorean triangles $a^2 + b^2 = p^2$ in which the hypotenuse $p$ is a prime number. Let $h(x) = \sum_{p \le x}p^2$, $a(x) = \sum_{p \le x}ab$ and $r(x) = \sum_{p \le x}(a+b)^2$. Is it true that:

$$ \lim_{x \to \infty}\frac{h(x)}{r(x)} = \frac{\pi}{2+\pi} $$

$$ \lim_{x \to \infty}\frac{a(x)}{r(x)} = \frac{1}{2+\pi} $$

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    $\begingroup$ you might want to give some evidence for these limits. $\endgroup$
    – Carlo Beenakker
    Commented Feb 12, 2022 at 8:52
  • $\begingroup$ @CarloBeenakker You are right. There was a typpo in the second limit. Corrected it. Here is the motivation for these two limits. mathoverflow.net/questions/348448/… $\endgroup$
    – Nilotpal Kanti Sinha
    Commented Feb 12, 2022 at 9:45

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Q: Is $\lim_{x \to \infty}\frac{h(x)}{r(x)} = \frac{\pi}{2+\pi} $ ?
A: use that $r(x)=h(x)+2a(x)$, hence $$\frac{h(x)}{r(x)} = \frac{h(x)/a(x)}{2+h(x)/a(x)}$$ and $$\lim_{x\rightarrow\infty}\frac{h(x)}{a(x)}=\pi$$ in view of https://math.stackexchange.com/a/3481801/87355

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