1

Let $b,c\in \mathbb{N}$ such that $b\gt c$. Let $r_{-1}=b$ and $r_0=c$. Then we define the numbers $q_i$ and $r_i$ as the quotient and remainder of euclidean division of $r_{i-2}$ and $r_{i-1}$, i.e. $$r_{i-2}=r_{i-1}q_i+r_i$$ where $0\le r_i\lt r_{i - 1}$. Let $j$ be the index of the last nonzero remainder. Then $r_{j+1}=0$ and $r_j=\gcd (b, c)$.

We will also define $$x_{-1}=1,\space\space\space x_0=0,\space\space\space x_i=x_{i-2}-q_ix_{i-1}$$ $$y_{-1}=0,\space\space\space y_0=1,\space\space\space y_i=y_{i-2}-q_iy_{i-1}$$ In these sequences, $bx_i+cy_i=r_i$ for all $i \in \{-1, 0, 1, ..., j, j + 1\}$ so $x_j$ and $y_j$ are Bézout's coefficients for $b$ and $c$.

A property that I have found to hold is that $\gcd (x_i, y_i)=1$ for all $i \in \{-1, 0, 1, ..., j, j + 1\}$. I haven't found a counterexample so I feel as though it holds. Does the statement hold, and what would a proof look like if it does? I'm quite frustrated because it seems like proving the statement should be trivial, and I'm sure that it is. Any insight and help are appreciated.

Natrium
  • 129
  • Correction: when I said coprime I meant that their $\gcd$ is $1$. – Natrium Aug 12 '22 at 16:16
  • 1
    I have a feeling it has nothing to do with the Division algorithm (the division algorithm determines the remainders and the factors of the remainders but the quotients are in essence arbitrary). I figure if you statement is true then it is probably true for any sequence of $q_i$. It seems we are in essence add coprime numbers to the same $q_i$ so the result is always coprime. – fleablood Aug 12 '22 at 16:17
  • Oh... I guess you are correct. Everything is coprime to $1$. I guess my statement should have been $0$ is only coprime to $\pm 1$. Sorry, my mistake. – fleablood Aug 12 '22 at 16:20

1 Answers1

3

Successive rows $\,(x_i,y_i),\, (x_{i+1},y_{i+1})\,$ form a matrix of determinant $1$, since they start as the identity matrix, and they update by elementary row operations (which preserve the determinant). Thus a common divisor of a row $\,\color{#c00}{x_i,y_i}$ divides the determinant $\,1 = \color{#c00}{x_i}\:\!y_{i+1}-x_{i+1}\color{#c00}{y_i},\,$ so $\,\gcd(\color{#c00}{x_i,y_i})=1\,$ as claimed.

Remark $ $ See this answer for more on the row operation view of the extended Euclidean algorithm (these ideas will become clearer when you study generalization of Gaussian elimination to modules, e.g. Hermite & Smith normal forms).

Bill Dubuque
  • 272,048