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Let us define recurrence equation as :

$S_n=S_{n-1}^{10}-10\cdot S^8_{n-1}+35\cdot S^6_{n-1}-50\cdot S^4_{n-1}+25\cdot S^2_{n-1}-2$ , with $S_0=12$

and let us define following notation :

$GF(n,20)=20^{2^n}+1$

Now , note that :

$GF(1,20) \mid S_1$ , $GF(2,20) \mid S_4$

$GF(1,20) \nmid S_2$ , $GF(1,20) \nmid S_3$ , $GF(1,20) \nmid S_4$

$GF(2,20) \nmid S_1$ , $GF(2,20) \nmid S_2$ , $GF(2,20) \nmid S_3$

$GF(3,20) \nmid S_1$ , $GF(3,20) \nmid S_2$ , $GF(3,20) \nmid S_3$

So ,one can formulate following assumption :

$GF(n,20)$ is prime iff : $GF(n,20) \mid S_{4^{n-1}}$ , where $n>0$

My question : What would be the easiest way to prove or disprove this statement ?

P.S.

Here you can find proof of correctness of Lucas-Lehmer primality test .

Mathematica code based on the assumption above (output is set of $GF(n,20)$ primes) :

Select[Table[GF = 20^(2^n) + 1; For[i = 1; s = 12, i <= 4^(n - 1), i++,  s = Mod[s^10 - 10*s^8 + 35*s^6 - 50*s^4 + 25*s^2 - 2, GF]]; If[s == 0, "GF" <> ToString@n, n], {n, Range[15]}], StringQ] 
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    I was also so confused a$t$ firs$t$...2012-01-30

0 Answers 0