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:

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

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

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:

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

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

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:

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

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

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:

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

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

Consider all Pythagorean triangles in which the hypotenuse is a prime number $\le x$. Let $h(x)$, $a(x)$ and $p(x)$ be the sum of the square of the hypotenuse, the sum of the areas and the sum of the perimeters all such triangles respectively. Is it true that:

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

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

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:

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

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