October 13th, 2015, 08:59 AM  #21 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
This guy uses the factorization of n1 https://en.wikipedia.org/wiki/Pockli...primality_test It is not ASTRONOMICAL for Charles. But Mobel talking about using it even if he was talking about something else and Charles start screaming it is ASTRONOMICAL!!! YOU NEED 10 BILLIONS OF DOLLARS while in his computer a primality could be checked in one second for a number od 300 digits. The same schemes like the cooks preparing dishes. You want this here is the recipe. I`m fed up. 
October 13th, 2015, 11:15 AM  #22 
Member Joined: Jun 2015 From: Ohio Posts: 99 Thanks: 19 
Mobel, did you even read the article you posted? https://en.wikipedia.org/wiki/Pockli...primality_test Finish the section that says example and you will find this quote. "The biggest difficulty with this test is the necessity of discovering prime factors of N  1, in essence, factoring N − 1. In practice this could be extremely difficult. Finding a_p's is a less difficult problem." Your own article proves that your method is difficult and contradicts what you said. 
October 13th, 2015, 11:50 AM  #23  
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41  Quote:
For sure no. Because I do NOT NEED to factorize n1. Read carefully first. I DO NOT NEED TO KNOW ALL THE PRIMES DIVIDING N1. ARE YOU BLIND!!!  
October 13th, 2015, 12:00 PM  #24 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
I gave the example of n=341 which can be quickly factorized. My method will work even with a number with 100000000000000 digits. I do not have the tools to present an example of 300 digits or more. Do not divert the discussion by saying " there is a method I can use in my computer which can tell you that a number of 100000 digits is a prime with certainty" Everybody know that the claim is false. Almost all the large primes used commercially are not 100% primes. So Charles please stop your lies. 
October 13th, 2015, 12:49 PM  #25 
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47 
Numberguru1 was addressing what you said about the link, not what you did in your methods. There is no need to offend and curse, also you are being ungrateful to the people who put their time through your conjectures in good will. Furthermore, it was said many times that given a too large $n$, it is obvious without the slightest question that $2^n$ will be very large, so even understanding that there is no direct factorization of $n1$ as you claim, because you are simply splitting it in each step as $\frac{n1}{2}$, then $\frac{n1}{4}$, etc, until reaching an odd power, in which case you can say "hey, these two numbers are factors of the original number", you are forming an analytical tree and there are two possibilities of interpretation; taking in your example with 341: 1) when you say that "either $\gcd(2^{170}+1,341)=341$ or $\gcd(2^{170}1,341)=341$" you mean to actually compute the gcd, in this case you are in a worst scenario than before, if $n$ is huge then $2^{n1}$ is MUCH larger than $n$, already for $n>3$ we have $2^{n1}>n$, so now you are let to factorize not $n$ but a series of astronomically larger power of 2; 2) when you say that "either $\gcd(2^{170}+1,341)=341$ or $\gcd(2^{170}1,341)=341$" you DO NOT mean to actually compute the gcd, you just mean that it is always one of the two cases and that one can go further below; if that's what was intended, you actually committed some errors, the first is a sort of petitio principii: 2.1) you don't know that $n$ is pseudoprime, or even a prime, beforehand, this is what you are testing, by no means e.g. $\gcd(2^\frac{n1}{2}+1,n)$ must be 1 (put $n=207$), or one is 1 and the other $n$ (put $n=161$), so if your number is a composite number which is not a pseudoprime your test is ruined because you cannot factor $2^\frac{n1}{2^x}+1$ as a difference of two squares in $\mathbb{Z}$; 2.2) suppose you are lucky, then you go to further steps but sadly $\gcd(2^\frac{n1}{2^x}1)=1$, so the observation above apply. 
October 13th, 2015, 01:11 PM  #26 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
I`m not ungrateful when someone read my post and give an answer. But one people do not read my post and answer to a question I did not even put in my post forcibly I will be angry. Computing gcd`s does not require factorization at all!!! So you are answering just for answering and that`s it. Same behaviour. Nothing have changed here. Waste of time. Ps : NO ONE IS REQUIRED TO ANSWER. IF THE POST DOES INTEREST YOU DO NOT READ IT. PUT ME ON IGNORE. I WILL NOT NOT DIE. I DO NOT EVEN CARE ABOUT BEING READ OR NOT. I PUT MY IDEAS PUBLICLY HERE AND ELSEWHERE. MY ONLY HOPE IS THAT SOMEONE WHO IS SMART ENOUGH CAN PUSH FORWARD MY IDEAS. 
October 13th, 2015, 01:26 PM  #27 
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47  
October 13th, 2015, 01:59 PM  #28 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41  
October 13th, 2015, 02:04 PM  #29 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
You said "1) when you say that "either gcd(2170+1,341)=341 or gcd(2170−1,341)=341" you mean to actually compute the gcd, in this case you are in a worst scenario than before, if n is huge then 2n−1 is MUCH larger than n, already for n>3 we have 2n−1>n, so now you are let to factorize not n but a series of astronomically larger power of 2;" YOU ARE WRONG!!! THERE IS NO NEED TO FACTORIZE ANYTHING. MORE THAN THAT I HAVE AN ALGORITHM BETTER TO COMPUTE GCD`S (PARTICULAR FOR THIS TEST AND ONLY FOR THIS PRIMALITY TEST). I WILL NEVER PUBLISH IT!!! I WILL KEEP IT JUST FOR ME!!! I WILL KEEP MUM UNTIL MY DEATH. MY BROTHER WILL INHERIT IT. 
October 13th, 2015, 02:05 PM  #30 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
And now good luck to your forum! A laughable forum built on one leg!!! Without charles the forum is dead!!! 

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