20170619, 21:10  #78  
"Curtis"
Feb 2005
Riverside, CA
11×461 Posts 
Quote:
There has been plenty of math shown in this thread for why your method (1) fails, and (2) is slower than trial division. There is nothing left for Batalov to demonstrate mathematically, so he chooses to point out that you're defending utter crap; you don't take the hint, so he's blunter than the rest of them. You should listen to him, really. 

20170619, 21:22  #79  
Aug 2006
175B_{16} Posts 
Quote:
Code:
sqrtup(n)=n=ceil(n); if(issquare(n,&n),n,sqrtint(n)+1) mahbelExtended(n)=my(a2,b2,c2,d,d2,g); for(a=sqrtup(n/4),sqrtint(n), a2=a^2; for(b=sqrtup((na2)/3),min(sqrtint(na2),a), b2=b^2; for(c=sqrtup((na2b2)/2),min(sqrtint(na2b2),b), c2=c^2; if(issquare(na2b2c2,&d)&&c>=d, d2=d^2; g=gcd(n,a); if(g>1 && g<n, return(g)); g=gcd(n,b); if(g>1 && g<n, return(g)); g=gcd(n,c); if(g>1 && g<n, return(g)); g=gcd(n,d); if(g>1 && g<n, return(g)); g=gcd(n,a2+b2); if(g>1 && g<n, return(g)); g=gcd(n,a2+c2); if(g>1 && g<n, return(g)); g=gcd(n,a2+d2); if(g>1 && g<n, return(g)); g=gcd(n,a+b); if(g>1 && g<n, return(g)); g=gcd(n,a+c); if(g>1 && g<n, return(g)); g=gcd(n,a+d); if(g>1 && g<n, return(g)); g=gcd(n,b+c); if(g>1 && g<n, return(g)); g=gcd(n,b+d); if(g>1 && g<n, return(g)); g=gcd(n,c+d); if(g>1 && g<n, return(g)); g=gcd(n,a+b+c); if(g>1 && g<n, return(g)); g=gcd(n,a+b+d); if(g>1 && g<n, return(g)); g=gcd(n,a+c+d); if(g>1 && g<n, return(g)); g=gcd(n,b+c+d); if(g>1 && g<n, return(g)); g=gcd(n,a+b+c+d); if(g>1 && g<n, return(g)))))); "could not factor"; My machine is busy with other work, but feel free to use this code to check for numbers this algorithm cannot factor. 

20170619, 21:31  #80 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
@mahbel one thing you'll find when coding is efficiency is key. Any steps that can be decreased should be. One thing from your original post that still bugs me, is you are considering values that aren't primes multiplied by N. Is there a reason I'm not seeing because you can generate say any value that is a square times N, by multiplying by the square root all of a,b,c,d for N. You'll find a lot of them out ahead of time this works for finding at least a few of those. for example 3^2+5^2+7^2+11^2 = 204 ; 816 = 4*204 = 6^2+10^2+14^2+22^2 so some values of the multiplier you had out front don't really add much new information.
Last fiddled with by science_man_88 on 20170619 at 21:38 
20170619, 21:37  #81 
Aug 2006
3×1,993 Posts 
Serge is one of the computational heavies around here, if he's not computing it it's surely because he doesn't think it's worth his time. There's no question he could code something up (and get big iron to run it!) on if he really wanted to.

20170619, 21:42  #82 
Feb 2017
Nowhere
140D_{16} Posts 

20170619, 21:50  #83 
Aug 2006
3·1,993 Posts 

20170619, 23:44  #84  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}×3×5 Posts 
Quote:


20170619, 23:49  #85 
Aug 2006
3·1,993 Posts 
Honestly I think it's pretty hard to decide what multipliers are useful in general. The OLF paper tries to motivate the choice with some theory, I think, but ends up choosing a multiplier based on numerical experiments IIRC. Maybe some people know more  Silverman, likely, and a number of experts we have on the forum  but there's a lot of voodoo and trialanderror.

20170620, 00:09  #86  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
so for example 612=12*51 and has 4 representations coming from 3*51=153 multiplying the bases by 2. Last fiddled with by science_man_88 on 20170620 at 00:10 

20170620, 00:12  #87  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
111100_{2} Posts 
Quote:


20170620, 00:15  #88  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20170620 at 00:27 

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