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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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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

Carlo Beenakker
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