## Mathematical experiments on the computer, Volume 105 (Pure by Ulf Grenander

By Ulf Grenander

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**Example text**

Thus g < q 2 1 . 20) imply the following: assuming6 q > p, then the residue systems xq mod p and yp mod q contain all integers [less than] p 1, each half of them. The number of residues in yp mod q is q 2 1 , hence this system contains q p residues larger than p. If G 0 among them are even and U 0 are odd, then we 2 have U 0 C G0 D q p : 2 Moreover we easily see that GD p 1 2 C G0; U D p 1 2 C U 0; which implies U G Á U 0 C G0 Á q p 2 mod 2: 6 Proof by Zeller [73] 1. According to GAUSS’s Lemma we have .

The exponents of the p 1 terms inside the brackets are just the integers 1; 2; : : : ; p 1 since g is a primitive root modulo C1 p. Since the signs alternate, we see that x g xg ˙ : : : D ˙G. The sign of G is that of . 1/p x, and since p is odd we conclude that ˙G D . 1/ G. g 2 / Á . pq / mod p, and since g p 1 2 Á 1 mod p, this implies . 25) 2. x / D 1 C x C x g C : : : C x g . x / is, because g is a primitive root modulo p, divisible by 1 x p , hence by p 1 x p . x / will be divisible by 11 xx if 1 x 1 xp 1 xp Á 0 mod : 1 x 1 x For a proof we have to distinguish two cases.

Pq / mod p, and since g p 1 2 Á 1 mod p, this implies . 25) 2. x / D 1 C x C x g C : : : C x g . x / is, because g is a primitive root modulo p, divisible by 1 x p , hence by p 1 x p . x / will be divisible by 11 xx if 1 x 1 xp 1 xp Á 0 mod : 1 x 1 x For a proof we have to distinguish two cases. (I) a n d p a r e c o p r i m e. x / is divisible by 11 xx . (II) a n d p a r e n o t c o p r i m e. x / p is divisible by 11 xx . Collecting everything and recalling that g 0 C 1, g C 1, . . , g p the numbers 2, 3, .