20150521, 13:58  #1 
Romulan Interpreter
Jun 2011
Thailand
2^{3}×19×61 Posts 
Stupid question reloaded
Assuming I take out my magic hat and pull out of it a long list of Fermat factors candidates. They are all between few digits and few million digits. Not interesting how my magic hat produced them, by sieving, guessing, dreaming them in the night, or else.
My question is what is the best way (fast, existent program, win32/64) to check if they are indeed factors of some Fermat numbers? (you know, squaring mod q, checking if it is 1, etc, but doing this fast and in somehow "organized" way, like pfgw or cllr are doing) 
20150521, 14:35  #2  
Jun 2003
2^{3}×607 Posts 
Quote:


20150521, 16:21  #3  
Nov 2003
1110100100100_{2} Posts 
Quote:
you would know that your last sentence is complete nonsense. Hint: among the unfactored Fermat composites, the smallest possible factor has at least 4096 bits. 

20150521, 19:00  #4  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
1000010101011_{2} Posts 
Quote:


20150521, 22:59  #5  
Aug 2005
Seattle, WA
2^{3}·11·19 Posts 
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20150522, 02:10  #6  
Romulan Interpreter
Jun 2011
Thailand
2^{3}·19·61 Posts 
Quote:
641 is a factor of a Fermat number and only have 3 digits. What NEW info YOUR comment brings, and how does it help me solve my problem? @axn: thanks for the fast response, the icon was not necessary, I know about pfgw and how to use it, but think like that: if you would ask me 1 year ago, before I was crunching for crus, what program is the best to find R/S primes, I would answer the same as you answered me, including the icon, but meantime I found out about better/faster ways, (i.e srXsievers and cllr) of which I had no idea at the time. That is why I was asking. One never knows. I have a real life who couldn't care less about the factors of Fermat numbers, so you see, I couldn't know. Moreover, I don't have a magic hat, in spite of what Mr. Silverman thinks, I am trying to learn. But not all people a geniuses as his majesty.. 

20150522, 04:28  #7 
Jun 2003
2^{3}·607 Posts 

20150522, 06:51  #8  
Romulan Interpreter
Jun 2011
Thailand
2^{3}·19·61 Posts 
Quote:
Which remark, by the way, is not only unproductive, but, if I am to nitpick, is totally on the weeds. The cofactors of F12 and F13, for example, are a C1133 and respective C2391, according with this site, and they are KNOWN composites, which means they have a factor smaller than their square root. Which square roots have 1133/2=567 and 2391/2=1196 digits, or respective (by multiplying with log2(10)), 1882 and respective 3972 bits. So, THERE ARE unknown Fermat factors smaller than 4096 bits, for sure... (gotcha ) Last fiddled with by LaurV on 20150522 at 06:54 

20150522, 07:54  #9  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3·3,529 Posts 
Quote:


20150522, 11:19  #10  
Nov 2003
2^{2}×5×373 Posts 
Quote:
of the problem/solution. 

20150522, 15:05  #11 
Romulan Interpreter
Jun 2011
Thailand
2^{3}·19·61 Posts 
Sure, you are right. I saw you wrote "among the unfactored composites" and I know you referred to the Fermats with no factors known yet, that is why I said (and underlined) "if I am to nitpick". But I just wanted to get back to you

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