Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1991. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 20)

Aleksander Grytczuk and Jaroslaw Grytczuk: On generators in multiplicative group of the Fieid Zp

- do 1(1 the set of all generators of the group PROOF OF LEMMA 1: Let R be the set of all quadratic residues |f< modulo p. Then we have that R^ is a subgroup of Z

p and since p=2q+l the group R has the order q. If b<£R then b is not P p a generator in Z . Since p C2q> 2 = Cp-l)

q = C-l>

q = —1Cmod p>, hence by Euler Theorem we get 2q«NR

p. On the other hand we have C2q) 2 = Cp-1)

2 = 1Cmod p) and therefore 2q has the order 2 and cannot be a generator in Z

p. But the group Z^ has exactly •p Cp-1) = <p C2q) = q-1 generators and therefore the set. NR

f

>S(2q> is the set of all generators in Z*

. P LEMMA 2. Let g be a generator of the group Z where p=4k+3. Then -g is also a generator in Z^. PROOF OF LEMMA 2: Let p=4k+3 and g be a generator in Z*. Then we have Z* = { g

k ; k=l,2,. . . .p-1 = 4k+2 } . By Euler theorem we have g 2 = -1 and therefore g

2

k*

1 — -1From last equality we get C 2.1> -g

2 = g

2

k*

3 It is easy to see that C2k+3, 4k+2=p-l>=l and therefore g

2k+

3 is a generator in Z* thus by C2.1) Lemma 2 follows. PROOF OF THE THEOREM. First we remark that if p=2q+l where p,q are odd primes then p=4m+3, because for p=4m+l the equality p=2q+i is imposible. Hence p=8k+3 or p=8k+7. If