20140412, 19:16  #1 
Apr 2014
5·17 Posts 
Pretty Fast Primality Test (for numbers = 3 mod 4)
So, I was actually trying to prove Sophie Germain conjecture by utilizing Wilson's theorem with respect to properties of Mersenne factors, but instead I found a pretty neat (and probably not exceedingly useful) consequence of the theorem.
I won't bother going through the entirety of the math, as it took me several hours of work to get the result, but essentially we can reduce the requirement that (n1)! = 1 mod n for primality into a product expansion in variable k that runs until (n4m+3)/4, m chosen sufficiently small for the number to test, and check that it's congruent to a power of 2 determined by m. Since numbers = 3 mod p have an m such that the upper bound on k = 0, we just find that m and then calculate the power of 2 the product expansion must be congruent to. Unfortunately, as you keep going, you have to keep the residues (or atleast the product of the residues) in memory, which will probably result in a ridiculously large number after awhile. Still, if you've already calculated the residue product up to that point, checking this single congruence is a primality certificate. Maybe an example would be helpful at this point. product(2*k^2+9(k+1);k=0 to (n11)/4, n>=11, n = 3 mod 4) = +/2^((n7)/4) mod n product(2*k^2+9(k+1);k=0 to (n15)/4, n>=15, n = 3 mod 4)*5 = +/2^((n+5)/4) mod n product(2*k^2+9(k+1);k=0 to (n19)/4, n>=19, n = 3 mod 4)*135 = +/2^((n+17)/4) mod n (Note: 135 = 27*5 precalc) So, at first we have a product chain with no precalculated residue, just take my word for it that this is the correct congruence. If you mess around with Wilson's theorem enough you'll arrive at it. Since upper bound on the first one is 0 when n = 11 (by design), it's enough to check that 9 = +/2 mod 11 to prove primality, which of course 9 = 2 mod 11 so 11 must be prime. To reach the second equation, we calculate 2*k^2 + 9(k+1) mod n for k = (n11)/4, the previous lower bound. This leaves us with 8 in the denominator of the first term and 4 in the second term, so we multiply both sides by 8 (increasing the power of 2 by 3) so that we have integers to deal with. Since we can get rid of the terms with n in them, this boils down to 11^22*9*(114) (abs because +/ on the RHS makes sign unnecessary) = 121  126 = 5. Add that to the product chain, reduce the upper bound of k, and increase the power of 2 respectively, and you have a congruency to test for the next number = 3 mod 4. If we're only interested in checking the simple case when upper bound k = 0, we know that the curve at k = 0 is always 9, so we multiply 9 by the precalculated residue product on LHS and see if it's congruent to +/ the right power of 2. Here's a plug and play version for n = 4m+11: product(2*k^2+9(k+1); k=0 to (n4m11)/4, n>=4m+11, n = 3 mod 4) * product((4(mj)+7)^218*(4(mj)+3);j=0 to m1, m >=1) = +/2^(n10) mod n, n >=11 With the convention that the empty product = 1. The associated congruence that acts a primality certificate for n = 4m+11 is then: 9*product((4(mj)+7)^218*(4(mj)+3);j=0 to m1, m >=1) = +/2^(n10) mod n The first few terms in that product sequence (starting with n = 15): 9*5, 9*27, 9*91, 9*187 And their product: 9*5, 9*135, 9*12285, 9*2297295 And for comparison using Wilson's (n1)! theorem you'd need: 87178291200, 6402373705728000, 1124000727777607680000, 403291461126605635584000000 ... A tiny bit more useful to be doing quadratic calculations instead of factorial, right? Looking at my method, I think you could remove the condition n = 3 mod 4 and apply it to Wilson's theorem as well, but since you can only guarantee (n1)/2 is an integer, you halve the effectiveness (at the result of double the applicable numbers). Likewise, increasing the modulus to 8 doubles the effectiveness but halves the applicable numbers. A neat trade off. It is interesting to note that there is not even one sequence on OEIS that contains 5,27,91. I always consider it a win to find useful sequences that haven't been documented yet. The other product chain is documented in slightly different form n*(2*n3) = 2*n^2 6n, mine's just offset a bit to start with n >= 11... If you wanted to you could change the offsets and get formulas for n >=3 and 7, but meh. So it changes Wilson's theorem by greatly reducing the size of the numbers being multiplied by playing with properties of modular arithmetic and limiting it to numbers = 3 mod 4 to reduce the impact further by dividing the number of terms to multiply by 4. Theoretically it should be possible to come up with a similar solution that only works for numbers = 7 mod 8, 15 mod 16, and so forth, with greatly reduced calculation. EDIT: Some more actual examples of primality certificates: 9*5 = +/ 2^5 mod 15, 0 != +/2 mod 15, 15 is not prime 9*5*27 = +/ 2^9 mod 19, 18 = 18 mod 19, 19 is prime 9*5*27*91 = +/ 2^13 mod 23, 4 = 4 mod 23, 23 is prime 9*5*27*91*187 = +/ 2^17 mod 27, 0 != +/ 14, 27 is not prime Hmm, interesting pattern. It might be enough to say that if the product = 0 mod n, the number is composite. Otherwise prime. And it looks like the calculation method also tends to end up positive, so there may not be a need for +/ except when n = 11, which makes sense when I look at the math that arrives there... Wouldn't matter if the previous conjecture I just made is true. We wouldn't even have to calculate 2^(n10) mod n... Wow this algorithm might actually be useful. Improving further if that conjecture were true, then instead of storing the very large product of the terms in the expansion, it would be sufficient to store each term in the expansion and then gradually expand it, so check 9 mod n, then 9*5 mod n, then 9*5*27 mod n, and if you reached the (m+1)th term in the product expansion without hitting a 0, 4m+11 is prime. Call me super excited guys. This sounds /feasible/. And I believe you could make a similar product expansion to account for the other class p = 1 mod 4 to work in tandem with this to cover all primes. Or break it down into 1, 3, 5, 7 mod 8 classes to improve the calculation speed. Last fiddled with by tapion64 on 20140412 at 20:00 
20140412, 20:22  #2 
Apr 2014
5×17 Posts 
The post is now set in stone. I hope I'm not wrong >.>

20140413, 17:52  #3 
Apr 2014
5·17 Posts 
I did some more work on this. Essentially, if n = 4m+11, then the highest term to add to the product chain is 16*m^216m5 < 16*m^2, so sqrt(16*m^216m5) < sqrt(16*m^2) = 4m < 4m+11, thus we can guarantee that if 4m+11 is prime, it won't appear in the product chain before you test it. This is necessary but not sufficient to prove the conjecture.
There is a possibility that perhaps n is not prime, but its factors simply haven't appeared in the product chain yet. It seems unlikely that this would happen, but I'm not sure how to prove it. Still, the number is guaranteed composite if it divides that product, and 'probably' prime if it does not. To take it to the deterministic stage without proving my conjecture, you'd then have to verify that the residue is congruent to 2^(n10) mod n, which is still a fairly inexpensive operation to prove primality. Using a divide and conquer approach requires log(m+1) stages of multiplication mod n, so this algorithm is basically O(log(n)^2*log(log(n))*log(log(log(n))) using FFT. Although that might only hold if you have atleast (m+1)/2 parallel streams. Still, seems fast. Last fiddled with by tapion64 on 20140413 at 18:22 
20140413, 21:14  #4 
Apr 2014
1010101_{2} Posts 
Well, I'm a loser. I spent all this time just assuming that the algorithm didn't work for n = 1 mod 4... It actually does! The only thing that changes is how you calculate m. Instead of m = (n11)/4, m = (n9)/4, the congruence test will still work. I'm also getting closer to proving that if n is composite, it's sufficient to say that it will divide the product of the sequence. I wonder if there's a way to come up with linear instead of quadratic product sequences that would have the same effect? They need to grow somewhat quickly to accumulate factors and be large enough to be divisible by n, which is guaranteed by the quadratic growth, but honestly I don't see why you couldn't do the same thing with the right linear sequence.
To highlight: This is now the first ever general purpose, parallelizable primality test. I will be working on writing CUDA code to benchmark, as soon as Nvidia approves my registered developer application. 
20140413, 21:42  #5  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}·1,877 Posts 
Quote:
If you won't post the actual math and algorithm then do us all a favor: DON'T POST. This is a *math* forum. I read what you did post. It is full of undefined variables, undefined expressions, and general gibberish. However, the little bit that does make sense tells us that the algorithm (if it works; we only have your word for it which is not sufficient for scientific discussion) is pretty much worthless. If it takes (n4m+3)/4 iterations, then it is PURELY EXPONENTIAL ALGORITHM. Furthermore, you say below immediately below: "runs until (n4m+3)/4, m chosen sufficiently small for the number to test" is totally WRONG, because as a function of n, the smaller m is, the LARGER (n4m+3)/4 becomes. This claim alone makes me suspect that you don't know what you doing. Prove that I am full of it. Post the math and the full algorithm. 

20140413, 21:46  #6  
"Bob Silverman"
Nov 2003
North of Boston
7508_{10} Posts 
This, I believe.
Quote:
run perfectly in parallel. Stop making grandiose claims that you can not back up. Post the math or STFU. Last fiddled with by R.D. Silverman on 20140413 at 21:47 

20140414, 04:21  #7  
"Dana Jacobsen"
Feb 2011
Bangkok, TH
3^{2}·101 Posts 
Quote:
ECPP: Valente's 1992 thesis describes distributed ECPP in quite a bit of detail. Morain has been running distributed ECPP for a long time. I believe the Franke, Kleinjung, and Wirth group also had a distributed vesion. Modern Primo runs on multiple cores and parallelizes quite nicely. I've debated adding an MPI layer to my code. There are numerous ways to find parallelism. APRCL is a great method, but it isn't something I'm very familiar with. Bosma's 1990 thesis has a short section about parallelization. AKS: Trivially easy to parallel or distribute. Each of the s tests require only three total variables (the loop number a, the polynomial modulus r, and the input n). A very slow and not currently useful algorithm compared to others, but so easy to parallelize. The BLS75 methods (n1, n+1, hybrids) should be easily parallelizible, given that on nontrivial numbers they are stuck factoring large numbers. I don't believe these are really used any more on numbers that require deep factoring, as ECPP and APRCL are far more suitable. However there are quite a few numbers that do work. Or are we discussing compositeness tests ala BPSW? 

20140414, 06:18  #8 
Apr 2014
5·17 Posts 
Wow, dude. Calm down. First, when I say it took several hours, I mean that it's all kind of jumbled up across text files and I haven't had the time to make it nice and neat yet. The math is there, but it's not a formal proof, it's just me tweaking and testing and retweaking. I wanted to prove the last conjecture that it's sufficient to only test if the congruence is 0 before I put it all together in a formal proof. I tried to give bounds and conditions on numbers where necessary, but like I said this isn't the formal proof yet. It should be obvious that all variables discussed here are integers, and their bounds are stated unless I missed a few.
Anyways, it's not particularly hard to get to my version using basic manipulation of Wilson's Theorem that n is prime iff (n1)! = 1 mod n. Step 1. You can write (n1)! as (n1)*(n2)*...*((n3)/2)*...*(1) Step 2. You can split this into +/(((n1)/2)!)^2. The sign is uncertain because if (n1)/2 is odd, the sign will be negative, and if it's even the sign will be positive. Previously I solved this by forcing the condition n = 3 mod 4, but I have a better way now. So to account for this, we have (1)^((n3)/2) * (((n1)/2)!)^2 = 1 mod n. Step 3. Now we take the square root. If n = 3 mod 4, we get ((n1)/2)! = +/ 1 mod n. If n = 1 mod 4, we get sqrt(1)*((n1)/2)! = +/ 1 mod n. Note that (n1)^2 = n^2  2n  1 = 1 mod n, so n1 is always a solution for sqrt(1), but since we have +/ on RHS, it doesn't matter and we can use ((n1)/2)! = +/ 1 for all n = 1 mod 2. Step 4. (n1)/2 * 2 = n1 = 1 mod n, but again we have +/ on RHS, so we can get rid of both of those terms, giving us ((n3)/2)! = +/ 2 mod n Step 5. Here's where it gets tricky. I had to write this out by example a few times to get it right, so showing an example would probably work best here too. Take n = 11, so ((113)/2)! = (4)! = 4*3*2. Then look at n = 13, so ((133)/2)! = (5)! = 5*4*3*2. Notice again the discrepancy in the number of terms based on n = 1 or 3 mod 4. I'll work it for 3 first, and then show it for 1. Set 2 to the side for a moment, and you have 4*3. If you have n = 15 instead, you have 6*5*4*3. We're going to split it into two products of length (n7)/4. We can write it as (n3)/2*...*(n+5)/4 multiplied by (n+1)/4*...*3. We can write these as products over variable k as product((n2k3)/2); k = 0 to (n11)/4, n > = 11) and product(k+3;k = 0 to (n11)/4). We could actually do this for n >= 3 or 7, but those are trivial and 11 sets up nicely. Notice that in the first product, we have 2 in the denominator, and there are (n7)/4 of them. Then remember that we set aside 2 earlier, so we have (n11)/4 of them in denominator. Multiply both sides by 2^(n11)/4 and that makes the formula n2k3, and our RHS is now +/2^(n7)/4. Note that we can now remove n from the first product to get 2k3, and both of our products are independent of n and run on the same index. Since we have (2k+3) and sign doesn't matter, we have (2k+3)*(k+3) = 2k^2+9(k+1). This gives us the first base case: product(2*k^2+9(k+1);k=0 to (n11)/4, n>=11, n = 3 mod 4) = +/2^((n7)/4) mod n Now, that's fine and good and all, but here's why I said if you choose n = 4m+11 small enough, which was bad wording on my part: If you choose n = 11 in that formula, you're doing a product from k = 0 to 0, i.e. a single calculation of the product at 0, which is trivially seen to always be 9. In the form above, it's not easy to see how this is useful, because it's based on n >= 11. But the product can be "pushed forward" by calculating the value of the formula at the current upper limit of k = (n11)/4, tacking it on as a precalculated part of the product (or storing, or whatever), moving the limit for k down to (n15)/4. Thus if you have already calculated all the terms in the product sequence that came before that, choosing n = 15 results in a k = 0 which means we multiply the product sequence by 9 and check if it's equal to +/2^((n+5)/4) mod n. I'm getting tired now, and I've explained all of the math that I didn't put in the original post, so you can see that everything flows to the point of where my original post begins explaining. Oh, except for the product calculation for n = 1 mod 4... But maybe you'll just take my word that it works out nicely for now, or work it yourself. I'll finish explaining tomorrow. As to why you can't follow my original post from that point, maybe it was difficult to understand if you didn't see the math that got there? Perhaps it will no longer be gibberish to you. But I can guarantee that it all logically flows. I'll clarify whatever you're having trouble following. I'll just end by acknowledging that yes, there are some other general purpose methods that run in parallel, but most are probabilistic or unproven, or simply too expensive to do when n gets large like trial factoring. I should have specified that I meant it was the first deterministic /feasible/ general purpose algorithm to run in parallel. Since it's a sequence of multiplications mod n that can be calculated independently (and a large amount of it could be cached to save time), and then divide and conquer using the power given to us by the GPU's threading to recombine mod n, it becomes a very usable primality test for large numbers (atleast, numbers that we can reasonably do FFT mod n on). The odds are also highly likely to come up with a product that n will divide if n is composite within the first few steps of divide and conquer, based on the relative size of the numbers. But some of it will be just luck, if the right factors aren't grouped together until the last multiplication, it's the same worst case as n actually being prime. Anyways, you don't have to believe my findings. But atleast try and understand what I have written before insulting me. One more thing. I do realize that this is essentially O(n) multiplication operations, so it's technically an exponential time algorithm. But because we can precalc, cache, reduce via modulo before and after each stage, and divide and conquer on GPU, it will have more like an O(log(N)) speed. Last fiddled with by tapion64 on 20140414 at 07:10 
20140414, 07:09  #9  
Apr 2014
5×17 Posts 
Quote:
EDIT: Looked into Modern Primo, the website only showed times for running CPU, 5000 digit number took over 9 hours on a 4 core / 8 thread 3.4 GHz i7. It doesn't even mention a GPU port on the site, where'd you find this? EDIT2: Also, I can't believe I forgot to mention this, but we can use this to quickly factor the composite number as well, back tracking its tree to show which values of the sequence were multiplied that yielded in a multiple of n. It would require precalculating the factors of the sequence, but it'd be reusable for any number. Last fiddled with by tapion64 on 20140414 at 07:39 

20140414, 07:46  #10  
Jun 2003
2^{2}·3^{2}·151 Posts 
Quote:


20140414, 09:00  #11  
Romulan Interpreter
"name field"
Jun 2011
Thailand
19×541 Posts 
Quote:


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