20200730, 08:22  #1 
May 2017
ITALY
1D1_{16} Posts 
Procedure for factorizing N into O (1) ...Not.
Let N be an odd number to factorize such that N = p * q with (p + q4) = 0 (mod 8) and (qp + 2) = 0 (mod 4)
then the derivative of sqrt[(2 + 3 x)^2 (2*((3*N1)/81)/3+1  x + x^2)]sqrt[(5 + 3 x)^2 (2*((3*N1)/81)/3+1  x + x^2)] that will be in the form 1/2*[(36x^3+117x^2Ax+B)/(sqrt((x^2x+2*((3*N1)/81)/3+1)*(5+3x)^2))+(36x^3+9x^2+Cx+D)/((sqrt((x^2x+2*((3*N1)/81)/3+1)*(2+3x)^2)))] doing this system (4*b+2)^2(2*a1)^2=N, 1/2*[(36*1^3+117*1^2(A+36*(a1))*1+B+60*(a1))/(sqrt((a^2a+47)*(5+3*1)^2))+(36*1^3+9*1^2+(C+36*(a1))*1+D+24*(a1))/((sqrt((a^2a+47)*(2+3*1)^2)))]=[3*(4*b+2)^2+3]/[4*b+2] that this is simplified (4*b+2)^2(2*a1)^2=N, [(36*1^3+117*1^2(A+36*(a1))*1+B+60*(a1))/(53*1)+(36*1^3+9*1^2+(C+36*(a1))*1+D+24*(a1))/(2+3*1)]=[3*(4*b+2)^2+3] are a and b p=4*b+2(2*a1) ed q=4*b+2+(2*a1) Example (4*b+2)^2(2*a1)^2=N, [(36*1^3+117*1^2(A+36*(a1))*1+B+60*(a1))/(53*1)+(36*1^3+9*1^2+(C+36*(a1))*1+D+24*(a1))/(2+3*1)]=[3*(4*b+2)^2+3] , A=956 , B=1435 , C=830 , D=560 , N=187 > a=2 ;b=3 >p=11 ;q=17 If N = 4 * G + 1 multiply by 3 and it becomes 4 * H + 3 If N = 4 * H + 3 and the algorithm does not work, multiply by 5 What do you think? 
20200730, 09:53  #2 
May 2017
ITALY
721_{8} Posts 
unfortunately it doesn't work. the correct formula is this
(4*b+2)^2(2*a1)^2=N, [(36*1^3+117*1^2(A+36*a*(a1)/2)*1+B+60*a*(a1)/2)/(53*1)+(36*1^3+9*1^2+(C+36*a*(a1)/2)*1+D+24*a*(a1)/2)/(2+3*1)]=[3*(4*b+2)^2+3] 
20200730, 09:56  #3 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
13155_{8} Posts 

20200730, 10:03  #4 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10100000011001_{2} Posts 

20200731, 08:30  #5 
May 2017
ITALY
721_{8} Posts 
English
Hi retina and xilman I found an algorithm that factorizes N=[4*b+2]^2(2*a1) in O(1) which only works if a<[sqrt[32*b+9]+1]/2 Now I'll explain it Code:
M=((N*31)/81)/3 Let's rewrite this (4*b+2)^2(2*a1)^2=N so (36*1^3+117*1^2((M+3)*36+20+36*(a1)*a/2)*1+(M*60+55)+60*(a1)*a/2)/24=24*b*(b+1) If we put a = 2 we notice that solve (36*1^3+117*1^2((M+3)*36+20+36*(21)*2/2)*1+(M*60+55)+60*(21)*2/2)/24>24*(b1)*(b1+1) , (4*b+2)^2(2*a1)^2=((M*3+1)*8+1)/3 b>1/8*(a^2a2) > a<[sqrt[32*b+9]+1]/2 Example N=667=(4*b+2)^2(2*a1)^2=[(4*b+2)(2*a1)]*[(4*b+2)+(2*a1)] M=((667*31)/81)/3=83 solve (36*1^3+117*1^2((83+3)*36+20+36*(21)*2/2)*1+(83*60+55)+60*(21)*2/2)/24>24*(b1)*(b1+1) b<7 > b=6 This is in O (1) assigning to a value of 2 if we know that q and p are of the same order, for example 100 digits we can assign a value of 98 digits so as to be less than true a Italian Ciao retina e xilman ho trovato un algoritmo che fattorizza N=[4*b+2]^2(2*a1) in O(1) che funziona solo se a<[sqrt[32*b+9]+1]/2 Ora te lo spiego M=((N*31)/81)/3 Riscriviamo questa (4*b+2)^2(2*a1)^2=N cosΓ¬ (36*1^3+117*1^2((M+3)*36+20+36*(a1)*a/2)*1+(M*60+55)+60*(a1)*a/2)/24=24*b*(b+1) Se mettiamo a=2 notiamo che solve (36*1^3+117*1^2((M+3)*36+20+36*(21)*2/2)*1+(M*60+55)+60*(21)*2/2)/24>24*(b1)*(b1+1) , (4*b+2)^2(2*a1)^2=((M*3+1)*8+1)/3 b>1/8*(a^2a2) > a<[sqrt[32*b+9]+1]/2 Esempio N=667=(4*b+2)^2(2*a1)^2=[(4*b+2)(2*a1)]*[(4*b+2)+(2*a1)] M=((667*31)/81)/3=83 solve (36*1^3+117*1^2((83+3)*36+20+36*(21)*2/2)*1+(83*60+55)+60*(21)*2/2)/24>24*(b1)*(b1+1) b<7 > b=6 Questo Γ¨ in O(1) assegnando ad a valore 2 se conosciamo che q e p sono dello stesso ordine ad esempio 100 cifre possiamo assegnare ad a un valore di 98 cifre in modo da essere minori del vero a 
20200731, 09:12  #6 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
5,741 Posts 

20200731, 16:52  #7 
May 2017
ITALY
3×5×31 Posts 

20200731, 17:22  #8 
6809 > 6502
"""""""""""""""""""
Aug 2003
101Γ103 Posts
5·11·157 Posts 
If you work in math nowadays you need to learn some programming. Also, if you work in Astronomy. And also in a number of other fields it can be needed.
Go learn Java or Python or some other language. Once you have one of your procedures coded up, try running it on a 200 digit number. 
20200731, 17:35  #9  
May 2017
ITALY
721_{8} Posts 
Quote:
could you give me a little help: does it work in O (1)? 

20200731, 17:58  #10  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
5·2,053 Posts 
Quote:
Your turn. Go learn some programming and some computer science and we will then take our turn. Last fiddled with by xilman on 20200731 at 18:14 

20200731, 18:04  #11 
May 2017
ITALY
3×5×31 Posts 
for O(1) I mean the number of cycles in programming.

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