Let n > 2 be an integer, b an integer, 1 < b < n, gcd(b, n) = 1, gcd(b-1,n) = 1, and let h be the multiplicative order of b (mod n).

Let k be an integer, 1 <= k < n.

Let r

_{i} = remainder of k*b^(i-1) mod n [0 < r

_{i} < n], i = 1 to h.

A) Prove that

for a positive integer m.

~~B) Prove that m = 1 if and only if n is a repunit to the base b, and also that one of the~~ r

_{i} ~~is equal to 1.~~
Remark: The thread title refers to the fact that the r

_{i} form a coset of the multiplicative group generated by b (mod n) in the multiplicative group of invertible residues (mod n).

**Part (B) as I stated it above is completely wrong.** I forgot to multiply a fraction by 1.

There is, alas, no connection to n being a repunit to the base b, or to any of the r

_{i} being 1, as shown by the example b = 10, k = 2, n = 4649 .

Foul-up corrected in followup post.