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 2021-08-28, 16:11 #1 Dr Sardonicus     Feb 2017 Nowhere 168D16 Posts Coset conundrum 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 ri = remainder of k*b^(i-1) mod n [0 < ri < n], i = 1 to h. A) Prove that $\sum_{i=1}^{h}r_{i}\;=\;m\times n$ 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 ri is equal to 1. Remark: The thread title refers to the fact that the ri 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 ri being 1, as shown by the example b = 10, k = 2, n = 4649 . Foul-up corrected in followup post. Last fiddled with by Dr Sardonicus on 2021-08-29 at 13:21 Reason: xingif stopy
2021-08-28, 21:28   #2
uau

Jan 2017

8A16 Posts

Quote:
 Originally Posted by Dr Sardonicus A) Prove that $\sum_{i=1}^{h}r_{i}\;=\;m\times n$ for a positive integer m.
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*(b-1)=0 mod n, and S must be 0 mod n.
Quote:
 B) Prove that m = 1 if and only if n is a repunit to the base b, and also that one of the ri is equal to 1.
The "and also that one of the ri is equal to 1" part seems ambiguous or wrong. For k != 1, m may or may not equal 1?

n = 1111, b = 10, k = 2: m = 1
n = 1111, b = 10, k = 21: m = 2

2021-08-29, 13:26   #3
Dr Sardonicus

Feb 2017
Nowhere

23·251 Posts

Quote:
 Originally Posted by uau The "and also that one of the ri is equal to 1" part seems ambiguous or wrong. For k != 1, m may or may not equal 1? n = 1111, b = 10, k = 2: m = 1 n = 1111, b = 10, k = 21: m = 2
Mea culpa. I made a huge blunder. Actual characterization of the multiplier m follows, proof left as exercise.

Conditions restated for ease of reference:
Quote:
 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.
(B) Let A = k*(bh - 1)/n, and sb(A) the sum of the base-b digits of A. Then m = sb(A)/(b-1).

Last fiddled with by Dr Sardonicus on 2021-08-29 at 13:29

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