[QUOTE=gophne;475141]Hi jnml
Using the algorithm (sagemath), I get positive result for M3, M5 & M17. I could not test beyound M29 (dividend) due to unknown run time). e.g. M3 (2^71) #1......a=2^71..................Divisor under test~127....[COLOR="Red"]n2[/COLOR] < what the hell is this? #2......b=(2^71)2.............[COLOR="Red"]n~125[/COLOR] #3......c=1/2*(b+1).............[COLOR="Red"]Target Congruant~63[/COLOR] #4......d=2^b1...................Dividend~42,535,295,865,117,307,932,921,825,928,971,026,431 #5.......e=d mod a..............Congruant~63 #6.......c==e.......................True~127(divisor) is Prime[/QUOTE] So after all your talk about n this, n that, your own scribbling show that you suddenly call n2 "n". Then of course, your "n+2" is n. Congratulations! You proved that all primes are 2PRP. What is even more impressive is that all composite Mersenne numbers are 2PRPs (even if you don't know it), so your test will just call them all prime. No false negatives! 
[QUOTE=Batalov;475223]So after all your talk about n this, n that, your own scribbling show that you suddenly call n2 "n". Then of course, your "n+2" is n.
Congratulations! You proved that all primes are 2PRP. What is even more impressive is that all composite Mersenne numbers are 2PRPs (even if you don't know it), so your test will just call them all prime. No false negatives![/QUOTE] Hi batalov That is how I coded the algorithm (in SAGE)....if you like I could(should?) have coded the algorithm as follows as well I did the above to put the "number" being tested as step #1. #1.......a=(2^71)2.........[COLOR="Red"]Modulo Dividend???[/COLOR]~125 (should be the dividend "seed" value to get "d", the real Modulo dividend as per the algorithm. #2.......b=2^71...............Modulo Divisor~(a+2=127)....NUMBER UNDER TEST #3.......c=(a+1)/2............Modulo Target Congruant~63 #4.......d=2^a1..............Modulo Divident~ As above #5.......e=d mod a..........Algorithm Congruant=63 #6.......c==e....................True~127(Divisor) is Prime To simplify further you could simply call the Divident seed [B]25[/B] in step #1. That would make the divisor in step #2, [B]27[/B]~(n+2). or (a+2) if you like. I only used mersenne notation because jnml was working with M[I]n[/I] notation, which would be more practical if very large numbers are to be considered. I am trying my best to be clear, but bear with me that I am trying to translate my code into words. I will try to type some actual code of a working page as I code the algorithm in SAGEMATH. But everybody would code differently depending on their expertise in SAGEMATH. My coding skills in SAGE are moderate. 
[QUOTE=Batalov;475223]Congratulations! You proved that all primes are 2PRP.
What is even more impressive is that all composite Mersenne numbers are 2PRPs (even if you don't know it), so your test will just call them all prime. No false negatives![/QUOTE] No true negatives, either  but hey, who needs all that negativity, it's harshing the holiday spirit, and stuff. 
[QUOTE="old TV cartoon"]That's why we called him Neutron! He is always so positive![/QUOTE]
[COLOR=Wheat].[/COLOR] 
[QUOTE=Batalov;475233]That's why we called him Neutron! He is always so positive![/QUOTE]
Umm... Wasn't that the Protons? At least one of the Electrons refused to play game. 
[QUOTE=Batalov;475223]So after all your talk about n this, n that, your own scribbling show that you suddenly call n2 "n". Then of course, your "n+2" is n.
Congratulations! You proved that all primes are 2PRP. What is even more impressive is that all composite Mersenne numbers are 2PRPs (even if you don't know it), so your test will just call them all prime. No false negatives![/QUOTE] hi batalov I do not understand what you are saying about 2PRP's ....are you running the algorithm? What are your results? I do not get all "composite mersenne numbers" to be false positives, on the contrary, e.g [B]2^91 & 2^151[/B], do not register as a false positives but as composite in the algorithm. Please post your results, you could even do the calculation on a calculator or Excel at this level of magnitude. 
SAGEMATH CODE for False Positives
Here is SAGEMATH code to check for false positive (prime) results (for the algorithm) up to 1,000,000. Should only take a few minutes to run.
[1] x=0 [2] for y in range(1,1000000,2): z= next_prime(y) a= y+2; #Modulo Divisor....Number under test b= 2^y1; #Modulo Dividend c= b%a; #Algorithm Modulo Congruent d= (y+1)/2; #Target Congruent as per Algorithm if c==d: if z>a: x=x+1 print(a,x); # 'a' is the False Positive identified, 'x' is a counter The SAGEMATH software will automatically do the indentation for the code. 
Restatement of Algorithm for Use and Testing
If (2^n1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME, else COMPOSITE, [I]n[/I] being an element of the set of positive integers =>1

[QUOTE=gophne;475261]If (2^n1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME, else COMPOSITE, [I]n[/I] being an element of the set of positive integers =>1[/QUOTE]
Yeah, we heard you the first millionandsix times. Try (n+2) = 2047 = 23*89 in your formula. 
[QUOTE=gophne;475252]
I do not understand what you are saying about 2PRP's [/QUOTE] A person who walks off a 1km cliff also doesn't understand that he is already dead. But he is. [QUOTE=gophne;475261][B]Restatement of Algorithm for Use and Testing [/B]If (2^n1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME, else COMPOSITE, [I]n[/I] being an element of the set of positive integers =>1[/QUOTE] This is exactly what is called a 2PRP test. 1) It is >350 years old. 2) It is false. Try n+2 == 341. (2^n1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME? Wrong. It is composite. [URL="https://en.wikipedia.org/wiki/Fermat%27s_little_theorem#Pseudoprimes"]F. Sarrus in 1820 found 341[/URL] as one of the first pseudoprimes, to base 2. That was 197 years ago. Try n+2 == 561. (2^n1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME? Wrong. It is composite. Try n+2 == 645. (2^n1) [B]mod[/B] (n+2) = (n+1)/2, then (n+2) is PRIME? Wrong. It is composite. 
[QUOTE=Batalov;475266]A person who walks off a 1km cliff also doesn't understand that he is already dead.[/QUOTE]
It's not the fall which kills them. It's the hard stop at the end... 
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