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Gerry science forum beginner
Joined: 11 Nov 2005
Posts: 23
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Posted: Tue Jul 18, 2006 10:45 am Post subject:
Primality & Factoring
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Two questions relating the following function:
N(r,x,y,z)=r^2+(z^2y+(2z+1)x)r+z^2(z^2-1)y^2/4+z(z^2+(z-1)/2)yx+z(z+1)x^2
1) How can i determine solutions for variables r,x,y,z if N is given?
2) For which range of the variables r,x,y,z can N be prime?
Any comments are welcome. |
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Gerry science forum beginner
Joined: 11 Nov 2005
Posts: 23
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Posted: Tue Jul 18, 2006 12:57 pm Post subject:
Re: Primality & Factoring
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For example
493=N(r,x,y,z)=N(5,4,4,2)=N(9,8,4,1)=N(11,6,6,1)=N(17,12,8,0)
( Variables r,x,y,z >0 and integer are of interest) |
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Dave Rusin science forum Guru
Joined: 25 Mar 2005
Posts: 487
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Posted: Tue Jul 18, 2006 5:26 pm Post subject:
Re: Primality & Factoring
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In article <1153219509.405937.291020@i42g2000cwa.googlegroups.com>,
Gerry <GerryMrt@gmail.com> wrote:
| Quote: | Two questions relating the following function:
N(r,x,y,z)=r^2+(z^2y+(2z+1)x)r+z^2(z^2-1)y^2/4+z(z^2+(z-1)/2)yx+z(z+1)x^2
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i.e. 4N = (2r + z^2y+2xz+x )^2 -(x + yz)^2
| Quote: | 1) How can i determine solutions for variables r,x,y,z if N is given?
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Factor N = pq, so 4N = u^2 - v^2 with u = p+q, v = p-q. Then for
any y,z let x = p - q - yz, r = (p+q - (x+2xz+yz^2))/2 (which
turns out to be integral).
| Quote: | 2) For which range of the variables r,x,y,z can N be prime?
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This requires q to be in {1, -1, N, -N}. For each fixed q the "range"
is parameterized by the (y,z) pairs as above.
| Quote: | Any comments are welcome.
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Wrong newsgroup. Try sci.math.
dave |
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Gerry science forum beginner
Joined: 11 Nov 2005
Posts: 23
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Posted: Thu Jul 20, 2006 4:12 pm Post subject:
Re: Primality & Factoring
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1) Interesting approach.
Would it be possible to create a faster algorithm than normal factoring
by trying to fit p and q as follows:
N=pq, p=yz^2/2+(x-y/2)z+r , q=y(z+1)^2/2+(x-y/2)(z+1)+r
Gerry
Sorry about the wrong group. |
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