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]