Question

Product of incongruent primitive roots modulo a prime p is congruent to 1 modulo p\

Answer 1

Answer:

Step-by-step explanation:

Let m ∈ N and a ∈ Z satisfy (a, m) = 1. Then the order h of

a modulo m exists, and h | φ(m). Moreover, whenever a

k ≡ 1 (mod m), one

has h | k.

Proof. By Euler’s theorem, one has a

φ(m) ≡ 1 (mod m), and so the order of

a modulo m clearly exists. Suppose then that h is the order of a modulo m,

and further that a

k ≡ 1 (mod m). Then it follows from the division algorithm

that there exist integers q and r with k = hq + r and 0 6 r < h. But then we

obtain

a

k = (a

h

)

q

a

r ≡ a

r ≡ 1 (mod m),

whence r = 0. Thus we have h | k, and in particular we deduce that h |

φ(m).

Answer 2

Answer:

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Step-by-step explanation:

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