20031221, 10:57  #1 
Nov 2002
74_{10} Posts 
primility testing
I hope somebody can help me:
I have a number ( approx. 1520 digits) and i try to trial factor it to p(1000)=7919 and start afterwards a fermat test ( a^p mod p=a) with the bases a=2,3,5,7. Can i then say that the tested number is prime??? 
20031221, 12:11  #2 
Dec 2003
Belgium
101_{8} Posts 
No, you can't, but you have a pretty good assumption.
If you want to use Fermat's little theorem as a primality test for a number p you must test all prime exponents smaller than p1. Pseudoprimes for each base aren't very rare, but combinations of a few bases (4 like you did) gives you a good idea... michael 
20031221, 12:18  #3 
Nov 2002
2×37 Posts 
i thought that carmichael numbers are really rare and can be excluded by doing a little trail factoring.

20031221, 12:27  #4 
Dec 2003
Belgium
5×13 Posts 
They are not very frequent, but there are still enough to not be sure about the primality of a number...
If i'm correct there are 3 numbers smaller than 100.000 that pass fermat's little theorem for base 2,3,5 and 7 that are composite: 29341 = 13x37x61 46657 = 13x37x97 75361 = 11x13x17x31 michael 
20031221, 20:49  #5 
Nov 2002
1001010_{2} Posts 
but these numbers are eliminated by trial factoring up to 7000!!!

20031221, 20:54  #6 
Dec 2003
Belgium
65_{10} Posts 
But these numbers are only 5 digits too, i wouldn't know how big the factors could become for 1520 digit carmichael numbers...
michael Last fiddled with by michael on 20031221 at 20:55 
20031221, 21:10  #7 
Dec 2003
Belgium
65_{10} Posts 
721801 = 601x1201, that's already a little bigger, let's see if i can find some pseudoprimes to base 2,3,5,7 with only factors bigger than 7000...
Also: A number n is a pseudoprime to the base b if b^n1 is congruent to 1 modulo n. A Carmichael number is a composite number n such that b^n1 is congruent to 1 modulo n for every b that is relatively prime to n. So we are talking about composite numbers that are pseudoprime to bases 2,3,5 and 7 ... this are not necessarily Carmichael numbers. michael 
20031222, 02:45  #8  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Re: primility testing
Quote:
If one claims, or wants to say, simply that "[a certain number] is prime", that implies that it can be proven, absolutely and without any possible doubt, that the number is 100% definitely a prime number. If, instead, one is referring to a number which has been proven by trial factoring to have no factors smaller than some limit (e.g., 7000 or even 7 trillion) and, in addition, passes the Fermat pseudoprime test for 4 (or even 4000) different prime bases, then one can correctly say that the number is a probable prime, but one cannot correctly say that the number is a "prime" with no qualifying adjective ahead of the word "prime". So it's always necessary to refer to the latter example as a "probable prime" or "pseudoprime". 

20031222, 09:42  #9 
Nov 2002
1001010_{2} Posts 
so which other methods could i use to determine that a number is 100% prime???

20031222, 12:49  #10  
"Sander"
Oct 2002
52.345322,5.52471
29×41 Posts 
Quote:


20031222, 13:23  #11 
Nov 2002
1001010_{2} Posts 
but i want to implement this methods in a program so i cant use a webpage!!

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