Re: Finding the Number of Pythagorean Triples below a bound

*To*: mathgroup at smc.vnet.net*Subject*: [mg68345] Re: Finding the Number of Pythagorean Triples below a bound*From*: titus_piezas at yahoo.com*Date*: Wed, 2 Aug 2006 05:23:58 -0400 (EDT)*References*: <eaeqa3$53v$1@smc.vnet.net><200607300848.EAA25171@smc.vnet.net> <eakfgm$rl6$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hello all, My thanks to Peter and Andrzej, as well as those who privately emailed me. To recall, the problem was counting the number of solutions to a bivariate polynomial equal to a square, Poly(a,b) = c^2 One form that interested me was the Pythagorean-like equation: a^2 + b^2 = c^2 + k for {a,b} a positive integer, 0<a<=b, and k any small integer. I was wondering about the density of solutions to this since I knew in the special case of k=0, let S(N) be the number of primitive solutions with c < N, then S(N)/N = 1/(2pi) as N -> inf. For k a squarefree integer, it is convenient that any solution is also primitive. I used a simple code that allowed me to find S(10^m) with m=1,2,3 for small values of k (for m=4 took my code more than 30 mins so I aborted it). The data is given below: Note: Values are total S(N) for *both* k & -k: k = 2 S(N) = 4, 30, 283 k = 3 S(N) = 3, 41, 410 k = 5 S(N) = 3, 43, 426 k = 6 S(N) = 3, 36, 351 Question: Does S(N)/N for these also converge? For example, for the particular case of k = -6, we have S(N) = 2, 20, 202 which looks suspiciously like the ratio might be converging. Anybody know of a code for this that can find m=4,5,6 in a reasonable amount of time? Yours, Titus

**Follow-Ups**:**Re: Re: Finding the Number of Pythagorean Triples below a bound***From:*danl@wolfram.com

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