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 2014-08-15, 05:33 #1 primus   Jul 2014 Montenegro 2·13 Posts Disproven Primality Test for Specific Class of kb^n-1 Definition : $\text{Let}$ $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)$ $, \text{where}$ $m$ $\text{and}$ $x$ $\text{are nonnegative integers .}$ Conjecture :$\text{Let}$ $N=k\cdot b^n-1$ $\text{such that}$ $n>2$ , $k$ $\text{is odd positive number ,}$ $3 \not\mid k$ , $b$ $\text{is even positive number ,}$ $3 \not\mid b ,$ $k $\text{Let}$ $S_i=P_b(S_{i-1})$ $\text{with}$ $S_0=P_{bk/2}(P_{b/2}(4))$ , $\text{thus}$ $N$ $\text{is prime iff}$ $S_{n-2} \equiv 0 \pmod{N}$ 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
2014-08-15, 12:40   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by primus Definition : $\text{Let}$ $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)$ $, \text{where}$ $m$ $\text{and}$ $x$ $\text{are nonnegative integers .}$ Conjecture :$\text{Let}$ $N=k\cdot b^n-1$ $\text{such that}$ $n>2$ , $k$ $\text{is odd positive number ,}$ $3 \not\mid k$ , $b$ $\text{is even positive number ,}$ $3 \not\mid b ,$ $k $\text{Let}$ $S_i=P_b(S_{i-1})$ $\text{with}$ $S_0=P_{bk/2}(P_{b/2}(4))$ , $\text{thus}$ $N$ $\text{is prime iff}$ $S_{n-2} \equiv 0 \pmod{N}$ 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
One question why use 4 for the general form but 3 for your b=6 form ? Also I don't see why this generalization couldn't be put in the other thread.

2014-08-15, 13:27   #3
primus

Jul 2014
Montenegro

2610 Posts

Quote:
 Originally Posted by science_man_88 One question why use 4 for the general form but 3 for your b=6 form ? Also I don't see why this generalization couldn't be put in the other thread.
I started a new thread because this conjecture uses 4 for starting value , not 3 , as you noticed , and that's why it couldn't be named as generalization of the conjecture for $k\cdot 6^n-1$ . This is rather some sort of generalization of the Riesel primality test .

 2014-08-16, 02:52 #4 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100101100010002 Posts 4103*10^7-1 is a counterexample. 41029999999 = 47743 · 859393. Passes the test. Ergo, your test is a PRP test. $\qed$
 2014-08-16, 02:58 #5 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23·1,201 Posts 68761*50^3-1 is another counterexample.
2014-08-16, 05:35   #6
primus

Jul 2014
Montenegro

2·13 Posts

Quote:
 Originally Posted by Batalov 4103*10^7-1 is a counterexample. 41029999999 = 47743 · 859393. Passes the test. Ergo, your test is a PRP test. $\qed$
Can you find counterexample for $5\not\mid b$ ?

2014-08-16, 06:02   #7
R.D. Silverman

Nov 2003

1D2416 Posts

Quote:
 Originally Posted by primus Can you find counterexample for $5\not\mid b$ ?
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.

2014-08-20, 17:43   #8
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

23·1,201 Posts

Quote:
 Originally Posted by primus Can you find counterexample for $5\not\mid b$ ?
[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 $5 \not\mid b$ 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, down-to-earth example how a very specific test becomes a proven prime test, read e.g. Berrizbeitia, Iskra, Math. Comp. 79 (2010), 1779-1791.

Or read about Konyagin-Pomerance test (e.g. in PN-ACP, 2006) -- why is this a valid test? Because all proper divisors > 1 are proven to be non-existent if the test passes.

 2014-08-21, 15:16 #9 primus   Jul 2014 Montenegro 2·13 Posts Thanks for help and references ! Last fiddled with by primus on 2014-08-21 at 15:36

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