20180121, 12:37  #1 
Dec 2010
74_{10} Posts 
Maximum number of prime factors (3 questions)
Hello, an amateur mathematician here...
Let's say we are interested in investigating a "random" large odd number we know next to nothing about like 1,990,292,873. We then begin applying trial division by a list of the primes in order to unlock some of its mysteries... Division by 3 yields a remainder of 2. Division by 5 yields a remainder of 3. Division by 7 yields a remainder of 2. Division by 11 yields a remainder of 8. Division by 13 yields a remainder of 10. So far, no luck in finding a factor... We then wonder what the maximum number of factors this number could have, given the information we've generated. Calculating the 8th root of 1,990,292,873 gives us ~14.5333 Given we've ruled out all prime factors less than the 8th root, we can conclude that our large random number has at most 7 prime factors. This is a new piece of information. Calculating the 7th root of 1,990,292,873 gives us ~21.30182 Division by 17 yields a remainder of 6. Division by 19 yields a remainder of 9. Again, no luck with factoring, but given we've ruled out all prime factors less than the 7th root, we can conclude that our large random number has at most 6 prime factors. This reduces the possibilities and, in my mind, represents greater information; a new level, if you will. The 6th root is ~35.466 The 5th root is ~72.4 The 4th root is ~211.2 The cubic root is ~1257.879 The square root is ~44612.7 So performing divisions by 23, 29, and 31 will limit the number to a maximum of 5 prime factors. Performing divisions by 37  71 will limit the number to a maximum of 4 prime factors. Performing divisions by 73  211 will limit the number to a maximum of 3 prime factors. Performing divisions by 223  1,249 will limit the number to a maximum of 2 prime factors (a semiprime). Finally, performing divisions by 1,259  44,587 would complete the factoring process. I now have a question. Are there any ways (or an improvement) of limiting the maximum number of prime factors of an integer aside from the above method of trial factoring up to the Nth root where N1 would be the new maximum number of prime factors? Personally, I've never seen this kind of question raised, and I know of no algorithms or methods aside from determining the actual factors (or primality) of the number. A second, related question I have is whether or not this question has been raised for the remaining cofactor of F(12) (the Fermat number 2^(2^12) + 1). I believe this cofactor is 1,133 digits long. If mathematicians can't find the next factor, perhaps they could at least determine what "level" they're on and focus on "leveling up" as they gain more information about the F(12) cofactor. Of course, I'm sure such factoring "levels" could be tracked for all sorts of particular numbers. A third, related question I have is whether or not it's somehow easier to complete the factorization of a number if a new limit on the maximum number of prime factors is reached. Alternatively, is such information of any use, or would it be nothing more than a fun way to track one's progress as they factor a random large integer? 
20180121, 12:58  #2 
"Forget I exist"
Jul 2009
Dumbassville
2^{4}×3×5^{2}×7 Posts 
CRT could get you a remainder class on some of these mod a primorial. Then maybe find out all you can about the factors of that remainder.

20180121, 13:02  #3 
Einyen
Dec 2003
Denmark
2^{3}×3^{2}×47 Posts 
Regarding F12 you can follow the ECM progress here (the top line 4096)
https://www.mersenne.org/report_ecm/ All 112,000 curves has been done at the 60 digit level B1=260M, which means the chance/risk of missing a factor at 60 digit is 11/e ~ 37%. But then 68,721 curves have been done on the 65 digit level B1=800M which lowers that chance greatly, so only a very small chance/risk that a 60 digit factor have been missed, and infinitesimal chance that an even smaller factor exist. If you want to run ECM on F12 using Prime95 here is the worktodo line: Code:
ECM2=1,2,4096,1,800000000,80000000000,1,"114689,26017793,63766529,190274191361,1256132134125569,568630647535356955169033410940867804839360742060818433" Last fiddled with by ATH on 20180121 at 13:03 
20180121, 14:32  #4  
Feb 2017
Nowhere
1760_{16} Posts 
Quote:
Another oldfashioned approach is to find quadratic residues of your number, since a quadratic residue mod N will be a quadratic residue mod each prime factor of N. Each residue eliminates about half the candidate trial factors. 

20180121, 18:55  #5  
Aug 2006
5,987 Posts 
Quote:
Trial division is the obvious approach to ruling out small factors. Strassen's algorithm is faster than trial division past some point if you need to guarantee that a number has no small prime factors. If you're OK with high probability (say, the chance of failure is at most one in a trillion trillion) you can use ECM very effectively. You can even use it deterministically, but only to relatively small bounds  Chelli shows that this can be done to 2^32. If you want to factor the number entirely, do a small amount of trial division, a bit of ECM, SQUFOF (or rho), a bunch more ECM, and then finish whatever is left with SIQS (< if 100 digits or so) or NFS. Quote:
\[\frac{0.016807\ldots\cdot0.0076\ldots}{1  0.016807\ldots + 0.016807\ldots\cdot0.0076\ldots} = 0.000128\ldots\] or about one in 7787. It's actually a bit less likely than that because the factor might have been found in t55 or lower. If you repeat the calculations for a 55digit factor, say, it would be significantly less likely. Quote:


20180122, 02:37  #6  
Dec 2010
2×37 Posts 
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As I've thought about this topic more I've noticed that if a number fails a Fermat test, it must have at least 2 prime factors. I also noticed that if a number is a Carmichael number, it must have at least 3 prime factors. Such observations just make me wonder if more could be said or discovered about the range of prime factors a number could be limited to. Perhaps investigation down this path could lead to something. I've also noticed that semiprimes have an interesting property. Assume N = p*q is a semiprime and that B and N are relatively prime. Then B^(N+1) must be congruent to B^(p+q) (mod N). I don't know how meaningful this is though. Last fiddled with by siegert81 on 20180122 at 02:39 Reason: Quotation was missing. 

20180122, 02:46  #7  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2,287 Posts 
Quote:
Thank you in advance. 

20180122, 05:08  #8  
Aug 2006
5,987 Posts 
Quote:
1, 11, 13, 17, 19, 25, 29, 41, 47, 53, 61, 79, 89, 107, 115, 121, 127, 139, 143, 149, 151, 155, 157, 167, 169, 179, 181, 185, 187, 193, 197, 209, 215, 221, 229, 247, 257, 275, 283, 289, 295, 307, 311, 317, 319, 323, 325, 335, 337, 347, 349, 353, 355, 361, 365, 377, 383, 389, 397, 415, 425, 443, 451, 457, 463, 475, 479, 485, 487, 491, 493, 503, 505, 515, 517, 521, 523, 529, 533, 545, 551, 557, 565, 583, 593, 611, 619, 625, 631, 643, 647, 653, 655, 659, 661, 671 (Also, potentially, the individual primes 2, 3, or 7.) So for example, knowing that 672 is a quadratic residue mod N proves that 5 does not divide N, while it's perfectly possible that 673 would divide N (since 673 = 1 mod 672 and 1 is a quadratic residue mod 672). Quote:
https://math.stackexchange.com/a/185657/1778 Various other algorithms that might be of interest, variants/extensions/alternatives: Hittmeir 2015 (see also talk) Hiary 2014 Rubinstein 2013/Kloosterman sums (blog entry) Costa & Harvey 2012 (see also their very accessible talk) Castagnos, Joux, Laguillaumie, & Nguyen 2009 Bostan, Gaudry, & Schost 2007 Boneh, Durfee, & HowgraveGraham 1999 McKee 1996 Lenstra 1984 Shanks/SQUFOF Lehman 1974 You'd have to decide which of these could work for you. Right. It did seem like the sort of thing you were looking for in terms of deciding a hard arithmetic property without factorization. Quote:
But with ECM it's not all that hard to take a number to t60. Quote:
I think this requires that N is a squarefree semiprime, p < q. Last fiddled with by CRGreathouse on 20180122 at 05:14 Reason: add BGS link 

20180122, 11:42  #9  
"Forget I exist"
Jul 2009
Dumbassville
2^{4}×3×5^{2}×7 Posts 
Quote:
B^(p+1) congruent to B^(p+1) mod q When I reduce. In the case p=q this gets us B^2 congruent to B^2 in both cases. 

20180122, 14:24  #10  
Feb 2017
Nowhere
2^{5}·11·17 Posts 
Quote:
As CRGreathouse also pointed out, if your quadratic residue R is too large to prepare a table of residue classes mod R in advance, you can quickly test R/p for each p. A traditional method of producing quadratic residues (mod N) is to develop the simple continued fraction for sqrt(N). This produces residues no larger than (if memory serves) 2*sqrt(N). BTW, PariGP factored your example number 1990292873 in a small fraction of a millisecond. It is the product of two primes. (Setting the debug level high enough reveals that Pollard rho disclosed one of the factors.) 

20180122, 16:00  #11 
Aug 2006
1763_{16} Posts 

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