20200507, 21:47  #12 
Mar 2016
3^{3}×11 Posts 
A peaceful and pleasant night for you,
if I regard only the primes p>=5 with p=x²+1 (x>1) and the primes p with p  (x²+1 ) with p > x can i derive any suggestion about the factorisation of p1 in advance (except divisible by 4 :) ) Some nice words toward this topic would be nice, the day was ugly for me. Greetings Bernhard 
20200508, 13:02  #13  
Feb 2017
Nowhere
3×7×199 Posts 
Quote:
There is one situation in which there is an easy algebraic factorization of p  1. If p = a^{2} + b^{2}, and b = k*a +/ 1 for some integer k, then p  1 = a*((k^{2} + 1)*a +/ 2*k). This includes p = x^{2} + 1 (k = 0), x^{2} + (x+1)^{2} (k = 1), etc. Unfortunately, such p are fairly thin on the ground... 

20200509, 18:56  #14  
Mar 2016
451_{8} Posts 
Quote:
if p=x²+1 then x  p1 if p  (x²+1) ??? Can I calculate q=x²+1 with q=r*p and x and r known; then f  p1 ; f ??? If someone knows a good answer would be very nice to get it. Greetings Bernhard Last fiddled with by bhelmes on 20200509 at 19:02 

20200516, 20:04  #15  
Mar 2016
3^{3}·11 Posts 
A peaceful and pleasant night for you,
Quote:
f(x)=x²+1 f(27)=10*73 Substitution with x=10k3 (k=3) gives f(k)=(10k3)²+1 = 100k²60k+10  Division by 10 = 10k²6k+1  1 since I need the factorisation of p1 =k(10k6) Therefore k=3 is a factor of p1 resp. 731 This is a good news and will give some results soon. Greetings Bernhard 

20200518, 12:42  #16  
Nov 2003
2^{2}·5·373 Posts 
Quote:
N, what transform one does when factoring f(N). Please explain your algorithm to find the transform without knowing the factorization of N. Note also that knowing 3  (p1) for pN does not help very much in practice. Finally, please explain the "good news". Stop giving yourself accolades. In point of fact, it is not "good news". It is just blind numerology. One day you may actually learn to listen to people who are experts. Such as post #13 https://www.mersenneforum.org/showpo...7&postcount=13 by Dr. Sardonicus. Failure to listen to/respect experts while asking for their advice is the sign of a fool. You never learn from what others try to teach you > you are unteachable. And I can't think of a worse insult. 

20200520, 20:07  #17 
Mar 2016
3^{3}×11 Posts 
A peaceful and pleasant night for you,
Perhaps I should explain the algorithm and the solution a little bit better. I have the factorisation of f(n)=n²+1=r*p where r element of N and p is prime I can assume that p is really a prime because of the construction of the quadratic sieve, detailled described under http://devalco.de/quadr_Sieb_x%5E2+1.php So, I want to find a non trival factor of p1 [4 (p1) ] From the point of the finishing I would like to have p1 = k(k+a) That means p=k(k+a)+1 I make a linear substitution in order to split the r from the quadratic equation, that means I substitute n=k*r+s where s=n mod r and s²+1 = 0 mod r Substitution gives f(k)=(kr+s)²+1 <=> k²r²+2krs+s²+1 s²+1 = r This is the trick after division by r I get p=k²r+2ks+1 1 p1=k²r+2ks p1=k (kr+2s) Goal reached Therefore I can state that k  p1 Perhaps someone understand this prove and perhaps someone will see the improvement. It is much better to know the factor k  p1 in advance especially if you want to check, if 2^[(p1)/k] = 1 mod p @Silverman: This is not numerology, but a nice piece of math. Have a pleasant night Bernhard 
20200520, 20:33  #18 
"Michael"
Aug 2006
Usually at home
2^{4}·5 Posts 
See 'Safe Primes' and Sophie Germain primes: https://en.wikipedia.org/wiki/Safe_prime

20200521, 02:21  #19  
Nov 2003
16444_{8} Posts 
Quote:
Once you know p you don't need to find a factor of p1. Or if you do then just factor p1. Quote:
a prime factor of p1. After all, p is now known. All the rest of the drivel given below is pointless Quote:
<bunch of irrelevant drivel deleted,.....> Quote:
Quote:
Yes, it would indeed elp the P1 factoring algorithm. But once again you failed to read Dr. Sardonicus' post that I referred to. Quote:
Once again you are giving yourself accolades for something that is just plain silly. Is your ego so weak that you have to tell the world that what you are doing is great? There is no way to know ( other than the factor of 4) a factor of p1 prior to factoring f(n). Your silliness starts by finding p! The problem is to find a factor of p1 without knowing p. Once you know p then just factor p1!! You start off by assuming that you already know what you are looking for. I'll say it again. You just don't listen. Will someone move this to misc.math? Last fiddled with by R.D. Silverman on 20200521 at 02:22 Reason: add request 

20200521, 05:57  #20  
Romulan Interpreter
Jun 2011
Thailand
2·23·199 Posts 
It looks like his goal is to factor p1, not to factor n or f(n). More exactly, he tries to factor a factor of p1 (call it x). For this he first "inflates" x to p, then p to f(n), and tries to do some "tricks" there. Unfortunately that will not work, as already pointed by the other posters.
Quote:
Find a prime p of the form mx+1 with natural m. Find a natural r such as y=rp=square+1 (this is what you call f(n)) Then you are in the right conditions to apply your algorithm. What is the substitution, beside of the trivial one (as you know p1=mx). Remember, you need to factor x. Last fiddled with by LaurV on 20200521 at 05:59 

20200521, 08:58  #21  
Mar 2016
3^{3}·11 Posts 
Quote:
A peaceful day for you, LaurV who should type a 20 digits example ? I will give a second example which is working: f(n)=n²+1 f(92)=5*1693 k=trunc (92/5)=18 1692/18=94 In the worst case k is equal 1, but there are a lot of numbers where k is a proper factor. @Silverman: You seem to be generous with some insults, LaurV with some jokes and me with some accolades. Sounds like chocolade and was a new word I have learned. Enjoy the day Bernhard Last fiddled with by bhelmes on 20200521 at 09:50 

20200521, 15:47  #22 
Romulan Interpreter
Jun 2011
Thailand
23C2_{16} Posts 

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