20220111, 14:58  #2025  
"James Heinrich"
May 2004
exNorthern Ontario
111000001100_{2} Posts 
George just found a pretty one (#16 biggest ever):
Quote:
Last fiddled with by James Heinrich on 20220111 at 15:06 

20220111, 20:22  #2026  
Feb 2017
Nowhere
2^{5}×167 Posts 
Quote:
M80309/10572519233/367135192227544403816033004684216729776734999 and M80309/10572519233 were the same. Is there some obvious reason for this? (My wits are presently addled by symptoms of a head cold...) 

20220111, 21:55  #2027  
"James Heinrich"
May 2004
exNorthern Ontario
7014_{8} Posts 
Quote:
Last fiddled with by James Heinrich on 20220111 at 21:56 

20220112, 13:34  #2028  
Feb 2017
Nowhere
2^{5}×167 Posts 
Quote:
The OP in that thread seems to say that when subsequent PRPCF tests say C, the new PRPCF residue replaces previous PRPCF residues. This would certainly account for all reported PRPCF residues being the same (assuming the remaining cofactor has tested composite). Why this would be done is beyond me, but the only alternative explanation that fits the facts seems to be that, as long as the remaining CF has tested composite, the original PRP residue is simply repeated. There may be good reasons for not publishing the sequence of actual PRP residues (mod 16^{16}) for the composite cofactors, of which I am ignorant. [I am rejecting the idea that the residues (mod 16^{16}) from the PRPCF tests are all actually the same.] Of course, if the remaining CF tests as a PRP, the "all PRP residues are the same" goes out the window, and the residue is reported as PRP_PRP_PRP_PRP_ . 

20220112, 14:54  #2029 
Jun 2003
3^{4}×5×13 Posts 
As I mentioned in that other thread, the residues produced are the same. You're assuming PRPCF does a standard Fermat test; it does NOT.
Let N=Mp/f Instead of checking 3^(N1)==1 (mod N) (equivalently 3^N==3 (mod N)), it computes 3^(N*f+1)=3^(Mp+1)==3^(f+1) (mod N) Note. 3^N==3 ==> 3^Nf == 3^f ==> 3^(Nf+1) == 3^(f+1) This gives rise to same residue, since we're always computing the same expression 3^(Mp+1). Advantages: 1) Each run produces same residue, hence multiple runs acts as additional checks on previous runs. 2) Since the modified computation is just a series of squarings, it is now amenable to GEC and CERT. 
20220112, 21:10  #2030  
Feb 2017
Nowhere
2^{5}×167 Posts 
Quote:
I had actually thought of the possibility that subsequent tests were simply looking at 3^(M + 1) (mod N) where N is the cofactor; N divides M = M_{p}. Suppose 3^(M+1) = M*q + R where 0 < R < M. Then, yes, N certainly divides 3^(M+1)  R = M*q = N*f*Q, so R may be considered to be "the residue" in that sense. However, what I usually think of as the "residue of 3^(M+1) (mod N)" is r, where 3^(M+1) = N*Q + r, and 0 < r < N. Clearly r is just R reduced mod N. Generally, r will be less than R. It was not clear to me why R  r would be divisible by 2^{64}. 

20220112, 21:43  #2031  
Jan 2021
California
3·101 Posts 
Quote:
ETA: This means that if the full residue of the PRP were saved, any time a (new) factor were found, the remaining cofactor could be checked against the original residue to see if it's PRP. Last fiddled with by slandrum on 20220112 at 21:48 Reason: Additional thought 

20220113, 00:58  #2032  
Feb 2017
Nowhere
2^{5}·167 Posts 
Quote:
Now 3^(M+1) = 3^(f*N + 1) = 3^(f*(N1) + f + 1) = (3^(N1))^{f} * 3^(f+1). So if 3^(N1) == 1 (mod N) we have (3^(N1))^{f} == 1 (mod N), and R == 3^(f+1) (mod N). Thus, if R =/= 3^(f+1) (mod N), the cofactor N is definitely composite. Done. No standard Fermat test needed. However, if R == 3^(f+1) (mod N) it does not follow that N is a base3 Fermat PRP, i.e. that 3^(N1) == 1 (mod N). Only that (3^(N1)))^{f} == 1 (mod N). I know, gcd(f, eulerphi(N)) would have to be greater than 1 in order for 3^(N1) not to be congruent to 1 (mod N). That seems extremely unlikely to me, and I am confident that no examples are known, but I don't know that it's impossible. 

20220113, 18:49  #2033 
Sep 2002
811 Posts 
P1 found a factor in stage #2, B1=766000, B2=25093000.
UID: Jwb52z/Clay, M108524239 has a factor: 952615068857130427852757781191 (P1, B1=766000, B2=25093000) 99.588 bits. 
20220115, 23:26  #2034 
"Vincent"
Apr 2010
Over the rainbow
AB4_{16} Posts 
M8590991 has a 121.408bit (37digit) factor: 3527086255292055773928440628536263153 (P1,B1=1560000,B2=627605550)
another big one. 
20220117, 15:06  #2035  
"James Heinrich"
May 2004
exNorthern Ontario
2^{2}×29×31 Posts 
Two nice firstfactor finds by anonymous:
Quote:
Last fiddled with by James Heinrich on 20220117 at 22:50 

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