20160907, 21:11  #23 
Aug 2006
2^{2}×5×293 Posts 
Here's a heuristic suggesting that my constant is right. Start by looking at the circle with radius sqrt(x). All lattice points in the circle are (u, v) with u^2 + v^2 < x. The circle has area \(\pi x\). Now look only at the first quadrant where u, v > 0, a quartercircle of area \(\pi x/4\). But we need to split it in half to get only the values with 0 < u < v which has area \(\pi x/8.\) Notice that the number of lattice points inside this area is equal to its area plus an error term equal to its outside (perimeter/circumference/etc.) for a total of \(\pi x/8+O(\sqrt x)\). So far this is a rigorous count of the number of points, but the next step is just a heuristic. The density of squarefree numbers is \(6/\pi^2,\) so the expected number of 0 < u < v with gcd(u, v) = 1 and u^2 + v^2 < x is about \(\pi x/8\cdot6/\pi^2 = \frac{3}{4\pi}x\) as desired.
The last step can probably be made rigorous  the area is pretty regular and the error bounds on squarefree density are small. 
20160908, 10:24  #24 
Mar 2016
FD_{16} Posts 
A peaceful day for you,
thank you for your efforts and explications. i know where the problem comes from: u should be odd and v even, or inverse ! so the tripple (u,v)=(1,3) is not mentionned in my list. i learned that for primitiv pyth. tripples that this fact is necessary, but i did not mentionned it, it was my fault. I will research if the amount of pyth. triples or the amount of different roots of 1 deliver a better approximation of pi. Nice greetings from the primes Bernhard 
20160908, 14:03  #25 
Aug 2006
13344_{8} Posts 
If you're going to arbitrarily reject some values then you can adapt my method and see if it comes out as desired.
