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Re: Finding the Number of Pythagorean Triples below a bound

• To: mathgroup at smc.vnet.net
• Subject: [mg68409] Re: Finding the Number of Pythagorean Triples below a bound
• From: titus_piezas at yahoo.com
• Date: Fri, 4 Aug 2006 03:59:22 -0400 (EDT)
• References: <eaeqa3$53v$1@smc.vnet.net><eakfgm$rl6$1@smc.vnet.net> <200608020923.FAA28520@smc.vnet.net> <79C36C70-E091-4A82-8EC5-0EDC743D081D@mimuw.edu.pl> <easiva$fp3$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

Hello Andrzej,

The code I used was this:

c1=Sqrt[N[a^2+b^2-k]];

countTriples[m_]:=Block[{a,b,i=0},For[a=1,a<(10^m/Sqrt[2]),a++,
For[b=a,b<10^m,b++,If[c1==Round[c1]&&c1<10^m,i++]]];i]

though m=3 takes about 50 sec (in an old comp) while m=4 will prolly be

around 5000 sec (83 mins). 


-Titus

• Follow-Ups:
  • Re: Re: Finding the Number of Pythagorean Triples below a bound
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>

• References:
  • Re: Finding the Number of Pythagorean Triples below a bound
    • From: titus_piezas@yahoo.com