20210828, 16:11  #1 
Feb 2017
Nowhere
2^{2}·3^{2}·139 Posts 
Coset conundrum
Let n > 2 be an integer, b an integer, 1 < b < n, gcd(b, n) = 1, gcd(b1,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^(i1) mod n [0 < r_{i} < n], i = 1 to h. A) Prove that for a positive integer m. 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 . Foulup corrected in followup post. Last fiddled with by Dr Sardonicus on 20210829 at 13:21 Reason: xingif stopy 
20210828, 21:28  #2  
Jan 2017
170_{8} Posts 
If you multiply the elements of the set {b^i for all i} by b, you get the same set with permuted elements. Thus multiplying the sum by b does not change it mod n. Thus S*b = S mod n, S*(b1)=0 mod n, and S must be 0 mod n.
Quote:
n = 1111, b = 10, k = 2: m = 1 n = 1111, b = 10, k = 21: m = 2 

20210829, 13:26  #3  
Feb 2017
Nowhere
1001110001100_{2} Posts 
Quote:
Conditions restated for ease of reference: Quote:
Last fiddled with by Dr Sardonicus on 20210829 at 13:29 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A "difference of two squares" conundrum  Dr Sardonicus  Puzzles  3  20190109 15:38 