Distinct solution to a linear congruence

Type: Article

Publication Date: 2010-10-16

Citations: 9

DOI: https://doi.org/10.2140/involve.2010.3.341

Abstract

Given n, k ∈ ‫ގ‬ and a 1 , a 2 , . . ., a k ∈ ‫ޚ‬ n , we give conditions for the equationA sufficient condition is that k ≤ φ(n) and a i be invertible in ‫ޚ‬ n for all i.If n > 2 is prime, the following conditions together are necessary and sufficient: k ≤ n, each a i is nonzero, and either k < n or not all of the a i are equal.Proof.We first construct y 1 , y 2 , . . .y k iteratively, as will be explained.For notational convenience, for i < j we set y i, j = y i (1 -a i+1 y i+1 )(1 -a i+2 y i+2 ) • • • (1 -a j-1 y j-1 ) MSC2000: 11B50, 11D79.

Locations

  • Involve a Journal of Mathematics - View - PDF
  • Project Euclid (Cornell University) - View - PDF

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