Re: Re: Finding the Number of Pythagorean Triples below a bound

*To*: mathgroup at smc.vnet.net*Subject*: [mg68381] Re: [mg68345] Re: Finding the Number of Pythagorean Triples below a bound*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 3 Aug 2006 06:07:01 -0400 (EDT)*References*: <eaeqa3$53v$1@smc.vnet.net><200607300848.EAA25171@smc.vnet.net> <eakfgm$rl6$1@smc.vnet.net> <200608020923.FAA28520@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 2 Aug 2006, at 11:23, titus_piezas at yahoo.com wrote: > 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 > Here is a piece code which utilises the ideas I have described in my previous posts: ls = Prime /@ Range[3, 10]; test[n_] := Not[MemberQ[JacobiSymbol[n, ls], -1]] && Element[Sqrt[n], Integers] f[P_, k_] := Sum[If[(w = a^2 + b^2 - k) < P^2 && test[w], 1, 0], {a, 1, P}, {b, a, Floor[Sqrt[P^2 - a^2]]}] g[m_, k_] := f[10^m, k] + f[10^m, -k] We can easily confirm the results of your computations,.e.g. for k=2. Table[g[i,2],{i,3}] {4,30,283} Since you have not revealed the "simple code" you have used it is hard to tell if the above one is any better. It is however, certainly capable of solving the problem for m=4: g[4,2]//Timing {4779.39 Second,2763} The time it took on my 1 Gigahertz PowerBook was over 70 minutes, which is longer than you thought "reasonable", so I am still not sure if this is any improvement on what you already have. The time complexity of this algorithm seems somewhat larger than exponential so I would expect that it will take about 6 hours to deal with n=5 on my computer, and perhaps 2 weeks to deal with n=6. Andrzej Kozlowski

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

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