20140815, 05:33  #1 
Jul 2014
Montenegro
2·13 Posts 
Disproven Primality Test for Specific Class of kb^n1
Definition : Conjecture : , , , Maxima implementation of the test
Maxima code to test this conjecture : Code:
/* input numbers b,k, b must be an even positive number not divisible by 3 , k must be an odd positive number not divisible by 3 , k<b^n */ k:5;b:10; for n from 3 thru 300 do (s:2*chebyshev_t(b*k/2,chebyshev_t(b/2,2)),N:k*b^n1, for i from 1 thru n2 do (s:mod(2*chebyshev_t(b,s/2),N)), (if((s=0) and not(primep(N))) then print(n)))$ 
20140815, 12:40  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20140815, 13:27  #3  
Jul 2014
Montenegro
2·13 Posts 
Quote:


20140816, 02:52  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×17^{3} Posts 
4103*10^71 is a counterexample. 41029999999 = 47743 · 859393.
Passes the test. Ergo, your test is a PRP test. 
20140816, 02:58  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·17^{3} Posts 
68761*50^31 is another counterexample.

20140816, 05:35  #6 
Jul 2014
Montenegro
26_{10} Posts 

20140816, 06:02  #7 
Nov 2003
2^{2}×5×373 Posts 
As with all mathematical results, reviews stop when the first error
is found. It is *your* job to perform a thorough review of your work before publishing it (presenting it here is a form of publishing). It is NOT the job of referees to find errors. Find your own counterexamples before asking others to do it for you. Responsibility for correctness of a result lies with the author, not the referees. 
20140820, 17:43  #8 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·17^{3} Posts 
[sarcasm]
Why? All you need to do is add a condition n>7 and your test will be all good and shining again. [/sarcasm] Frankly, though, no one will take either of these tightened conditions (either or n>7) seriously without an explanation why they are relevant. You are just fishing. As though if counterexamples didn't exist this conjecture would have been not wrong. For a simple, downtoearth example how a very specific test becomes a proven prime test, read e.g. Berrizbeitia, Iskra, Math. Comp. 79 (2010), 17791791. Or read about KonyaginPomerance test (e.g. in PNACP, 2006)  why is this a valid test? Because all proper divisors > 1 are proven to be nonexistent if the test passes. 
20140821, 15:16  #9 
Jul 2014
Montenegro
11010_{2} Posts 
Thanks for help and references !
Last fiddled with by primus on 20140821 at 15:36 
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