Finding the Number of Pythagorean Triples below a bound
- To: mathgroup at smc.vnet.net
- Subject: [mg68189] Finding the Number of Pythagorean Triples below a bound
- From: titus_piezas at yahoo.com
- Date: Sat, 29 Jul 2006 01:00:02 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Hello all,
Given the equation a^2+b^2 = c^2 with (a,b,c) positive integers,
0<a<=b. (For argument's sake, a=b will be allowed.) Let S(10^m) be the
number of solutions with hypotenuse c < bound 10^m. (They need not be
primitive.)
Problem: Find S(10^m).
What would be an efficient Mathematica code for this? The explicit
(a,b,c)'s are not needed, just S(10^m). I also need "m" as high as
possible, maybe m=6,7, with a reasonable run-time, say, less than an
hour. (A kind person gave me a code but told me it may be practical
only up to m=3.)
I stumbled upon some code some time ago, but it only gives the explicit
(a,b,c)'s with c<10^m, not the count:
Block[{a,b}, For[a=1,a<(10^m/Sqrt[2]),a++, For[b=a,b<10^m,b++,
If[Sqrt[a^2+b^2]==Round[Sqrt[a^2+b^2]]&&Sqrt[a^2+b^2]<10^m,
Print[{a,b,Sqrt[a^2+b^2]}]]]]]
for some positive integer "m". Can the above program be modified to
give S(10^m)? Or what program would be fast, for m=6,7?
Any help would be appreciated. Thanks.
-Titus
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- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
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