| American Citizen on Thu, 03 Jul 2025 05:14:14 +0200 |
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| Re: question on execution time for qfbsolve |
I ran over the first 100,000 integers ? for(i=1,100000,qfbsolve(Qfb(1,0,1),i,3)) cpu time = 3,394 ms, real time = 3,394 ms. ? for(i=1,100000,qfbsolve(Qfb(1,0,1),[i,factor(i)],3)) cpu time = 3,348 ms, real time = 3,348 ms. so it is about the same, not much improvement, 46 milliseconds. On 7/2/25 20:11, American Citizen wrote:
Hello:Currently I am using qfbsolve(Qfb(1,0,1),N,3) to find pairs of squares summing to N (if possible)The web page https://pari.math.u-bordeaux.fr/dochtml/html/Arithmetic_functions.html#qfbsolve states that "The integer n can also be given by its factorization matrix fa = factor(n) or by the pair [n, fa]."Currently my values for N < 1,000,000 and in fact are only primes or composites of primes of 1 mod 4 only.How much of a speed up would I obtain by doing qfbsolve(Qfb(1,0,1),[N,factor(N)],3) instead?Randall