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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
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