20200301, 17:52  #1 
"Sam"
Nov 2016
2^{2}·79 Posts 
Patterns in primes (no counterexamples)
Lemma: Let D = a^2+4 and p be a prime such that Jacobi(D  p) = 1.
Then (x^2 + a*x  1)^p = x^3  x^2 + (a^2 + 2)*x  a^2  5 mod(p, x^4+D*x^2+D) Or (x^2  a*x  1)^p = x^3  x^2 + (a^2 + 2)*x  a^2  5 mod(p, x^4+D*x^2+D) For instance, if D = 5, (a = ±1), then (a=1 if p = 3 mod 5) and (a = 1 if p = 2 mod 5). Also found were, D = 8, (a = ±2), then (a=2 if p = {11, 13} mod 16) and (a = 2 if p = {3, 5} mod 16) D = 13, (a = ±3), then (a=3 if p = {2, 5, 6} mod 13) and (a = 3 if p = {7, 8, 11} mod 13) D = 20, (a = ±4), then (a=4 if p = {3, 7, 13, 17} mod 40) and (a = 4 if p = {23, 27, 33, 37} mod 40) ... (and so on) The question here is what is the general pattern for determining which property will hold (the first one or second one)? What makes this test interesting is the primes in the same groups and residue classes mod D behave differently for this test. For instance, there is no polynomial which is irreducible for primes p = 2 mod 5 but not for primes p = 3 mod 5, or vice versa. Last fiddled with by carpetpool on 20200301 at 20:33 Reason: Typo error, (thanks to paulunderwood for noticing). 
20200301, 18:26  #2  
Sep 2002
Database er0rr
3,449 Posts 
Quote:
Code:
? a=5;D=a^2+4;p=107;Mod(Mod(1,p)*x^2a*x1,x^4+D*x^2+D)^p==x^3x^2+(a^2+2)*xa^25 0 ? a=5;D=a^2+4;p=107;Mod(Mod(1,p)*x^2+a*x1,x^4+D*x^2+D)^p==x^3x^2+(a^2+2)*xa^25 0 ? kronecker(a,p) 1 

20200301, 20:43  #3  
"Sam"
Nov 2016
2^{2}×79 Posts 
Quote:
Code:
(12:34) gp > kronecker(D,p) %43 = 1 (12:34) gp > a = 6 is quite interesting: "plus": p = {7, 11, 19, 21, 29, 33, 57, 63} mod 80 "minus": p = {17, 23, 47, 51, 59, 61, 69, 73} mod 80 Still, no pattern, and why does it appear to be so random? Last fiddled with by carpetpool on 20200301 at 20:48 

20200303, 00:34  #4 
"Sam"
Nov 2016
2^{2}·79 Posts 
I was curious to find composites passing one of these two tests. To my suprise, not a single [p,a] pair with (p, a) < 100k!
Code:
(21:18) gp > for(p=1,100000, if(isprime(p)==0, for(a=1,p, D=a^2+4; if(kronecker(D,p)==(1) & Mod(Mod(1,p)*x^2a*x1,x^4+D*x^2+D)^p==x^3x^2+(a^2+2)*xa^25, print([p,a]))))) (11:29) gp > Code:
(21:18) gp > for(p=1,100000, if(isprime(p)==0, for(a=1,p, D=a^2+4; if(kronecker(D,p)==(1) & Mod(Mod(1,p)*x^2+a*x1,x^4+D*x^2+D)^p==x^3x^2+(a^2+2)*xa^25, print([p,a]))))) (11:24) gp > 
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