PRIME CASE, p < 400, even k | p-1, k < p-1.
  LINK = some z is linked to 1 (Theorem 2.4 applies);  QUAD = p > (k-1)(k-2)+2 (Theorem 3.4);  BBC = E is 2-maximal
  p= 5 k= 2  LINK=1 QUAD=1 BBC=1
  p= 7 k= 2  LINK=1 QUAD=1 BBC=1
  p=11 k= 2  LINK=1 QUAD=1 BBC=1
  p=13 k= 2  LINK=1 QUAD=1 BBC=1
  p=13 k= 4  LINK=1 QUAD=1 BBC=0
  p=13 k= 6  LINK=0 QUAD=0 BBC=0
  p=17 k= 2  LINK=1 QUAD=1 BBC=1
  p=17 k= 4  LINK=1 QUAD=1 BBC=1
  p=17 k= 8  LINK=0 QUAD=0 BBC=0
  p=19 k= 2  LINK=1 QUAD=1 BBC=1
  p=19 k= 6  LINK=1 QUAD=0 BBC=0
  p=23 k= 2  LINK=1 QUAD=1 BBC=1
  p=29 k= 2  LINK=1 QUAD=1 BBC=1
  p=29 k= 4  LINK=1 QUAD=1 BBC=1
  p=29 k=14  LINK=0 QUAD=0 BBC=0
  p=31 k= 2  LINK=1 QUAD=1 BBC=1
  p=31 k= 6  LINK=1 QUAD=1 BBC=0
  p=31 k=10  LINK=0 QUAD=0 BBC=0
  p=37 k= 2  LINK=1 QUAD=1 BBC=1
  p=37 k= 4  LINK=1 QUAD=1 BBC=1
  p=37 k= 6  LINK=1 QUAD=1 BBC=1
  p=37 k=12  LINK=0 QUAD=0 BBC=0
  p=37 k=18  LINK=0 QUAD=0 BBC=0
  p=41 k= 2  LINK=1 QUAD=1 BBC=1
  p=41 k= 4  LINK=1 QUAD=1 BBC=1
  p=41 k= 8  LINK=1 QUAD=0 BBC=0
  p=41 k=10  LINK=0 QUAD=0 BBC=0
  p=41 k=20  LINK=0 QUAD=0 BBC=0
  p=43 k= 2  LINK=1 QUAD=1 BBC=1
  p=43 k= 6  LINK=1 QUAD=1 BBC=1
  p=43 k=14  LINK=0 QUAD=0 BBC=0
  p=47 k= 2  LINK=1 QUAD=1 BBC=1
  p=53 k= 2  LINK=1 QUAD=1 BBC=1
  p=53 k= 4  LINK=1 QUAD=1 BBC=1
  p=53 k=26  LINK=0 QUAD=0 BBC=0
  p=59 k= 2  LINK=1 QUAD=1 BBC=1
  p=61 k= 2  LINK=1 QUAD=1 BBC=1
  p=61 k= 4  LINK=1 QUAD=1 BBC=1
  p=61 k= 6  LINK=1 QUAD=1 BBC=1
  p=61 k=10  LINK=1 QUAD=0 BBC=0
  p=61 k=12  LINK=0 QUAD=0 BBC=0
  p=61 k=20  LINK=0 QUAD=0 BBC=0
  p=61 k=30  LINK=0 QUAD=0 BBC=0
TOTAL pairs (p,k): 353;  LINK: 231;  QUAD: 195;  BBC: 159;  Lemma 3.1 checked, max bad/bound = 1.000

PRIME POWERS p^n, n >= 2, E of order k | p-1: is p linked to 1 ?
  N=   9 k= 2  linked(1,p)=True
  N=  27 k= 2  linked(1,p)=True
  N=  25 k= 2  linked(1,p)=True
  N=  25 k= 4  linked(1,p)=True
  N= 125 k= 2  linked(1,p)=True
  N= 125 k= 4  linked(1,p)=True
  N=  49 k= 2  linked(1,p)=True
  N=  49 k= 6  linked(1,p)=True
  N= 343 k= 2  linked(1,p)=True
  N= 343 k= 6  linked(1,p)=True
  N= 121 k= 2  linked(1,p)=True
  N= 121 k=10  linked(1,p)=True
  N=1331 k= 2  linked(1,p)=True
  N=1331 k=10  linked(1,p)=True
  N= 169 k= 2  linked(1,p)=True
  N= 169 k= 4  linked(1,p)=True
  N= 169 k= 6  linked(1,p)=True
  N= 169 k=12  linked(1,p)=True
  N=2197 k= 2  linked(1,p)=True
  N=2197 k= 4  linked(1,p)=True
  N=2197 k= 6  linked(1,p)=True
  N=2197 k=12  linked(1,p)=True
  N= 289 k= 2  linked(1,p)=True
  N= 289 k= 4  linked(1,p)=True
  N= 289 k= 8  linked(1,p)=True
  N= 289 k=16  linked(1,p)=True
  N= 361 k= 2  linked(1,p)=True
  N= 361 k= 6  linked(1,p)=True
  N= 361 k=18  linked(1,p)=True
  N= 529 k= 2  linked(1,p)=True
  N= 529 k=22  linked(1,p)=True

COPRIME PRODUCTS N = N1*N2, E injective mod N1 and mod N2: are e1, e2 linked ?
  N=65 E=[1, 8, 57, 64]  linked(e1,e2)=True
  N=65 E=[1, 18, 47, 64]  linked(e1,e2)=True
  N=91 E=[1, 9, 10, 81, 82, 90]  linked(e1,e2)=True
  N=91 E=[1, 16, 17, 74, 75, 90]  linked(e1,e2)=True
  N=85 E=[1, 13, 72, 84]  linked(e1,e2)=True
  N=85 E=[1, 38, 47, 84]  linked(e1,e2)=True
  N=221 E=[1, 21, 200, 220]  linked(e1,e2)=True
  N=221 E=[1, 47, 174, 220]  linked(e1,e2)=True

GENERAL CASE (Lemma 3.1, Theorems 3.2 and 3.4): N < 130, all the groups -1 in E generated by at most 3 elements,
  m = max gcd(a-1,N) over a != 1 in E.  Checks: non-regular characters <= (k-1)(m-1)+1, and for any regular y,
  the number of z not linked to y is <= mk(k-1)-2(k-2).
  groups checked: 512, max bad/bound = 1.000, Theorem 3.4 applies to 183 of them, among which 81 with m > 1