20180130, 13:43  #1 
May 2017
ITALY
111111100_{2} Posts 
Semiprime factorization conjecture
Let N be a semiprimo then
N=a^2b^2 will have two solutions a1=(N+1)/2 and b1=(N1)/2 a2=? b2=? bruteforce on b2 starting from 0 A+B=a1 , AB=a2 , C+D=b1 ,CD=1 gcd(A,C)=p_a gcd(B,D)=p_b p=(p_a^2p_b^2) gcd(A,D)=q_a gcd(B,C)=q_b q=(q_b^2q_a^2) Example 15=a^2b^2 a=4 b=1 a=8 b=7 A+B=8 , AB=4 , C+D=7 , CD=1 A=6 ,B=2 ,C=4 ,D=3 gcd(A,C)=2 gcd(B,D)=1 p=(2^21^2)=3 gcd(A,D)=3 gcd(B,C)=2 q=(3^22^2)=5 What do you think about it? 
20180130, 13:59  #2 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

20180130, 14:03  #3 
May 2017
ITALY
2^{2}·127 Posts 

20180130, 14:53  #4 
May 2017
ITALY
111111100_{2} Posts 

20180203, 01:30  #5 
"Jane Sullivan"
Jan 2011
Beckenham, UK
3^{2}×31 Posts 

20180203, 02:07  #6  
"Sam"
Nov 2016
2^{3}×41 Posts 
Quote:
b = 1 (mod p) and 1 (mod q) c = 1 (mod p) and 1 (mod q) Your "conjecture" follows from this fact. Last fiddled with by carpetpool on 20180203 at 02:08 

20180203, 02:41  #7  
"Sam"
Nov 2016
101001000_{2} Posts 
Quote:
Here we have semiprimes N = p*q ranging from 90100 digits. Show that for each one, p = q = 1 (mod n) for a specific n. That is, show each prime dividing N has the form k*n+1. For example, N = 5539113502033412895292115974558480983800647735117931750794894952942621260596968647983584706901 (n = 90) is a semiprime. Show each prime dividing N has the form 90*k+1, or equivalently p = q = 1 (mod 90) where pq = N. Here are a few more examples (try and show that each prime dividing N has the form k*n+1): N = 448089051076275785687389418381162972987229036034444427278869256690561568953546204700396224357, (n = 36) N = 12875011596982825641079225756329235926774793871954944034524736059975902131902515285746205781, (n = 42) N = 325660878359113486502421634634373166016783555791941517818588883372438081215159258342011023201, (n = 100) N = 2529326286788100617652188740443724847554819001546212172410115864297421395751726116717093396417, (n = 98) 

20180216, 08:27  #8 
Aug 2006
3·1,993 Posts 
Nice game! Could you show us how it works?

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