3

If $m=a_1x+b_1y$, $n=a_2x+b_2y$ and $a_1b_2-a_2b_1=1$, then prove that $\mathrm{GCD}(m,n)=\mathrm{GCD}(x,y)$

May I get a hint, please?

Mathejunior
  • 3,344
  • 1
    By definition of gcd, if $m = a_1x + b_1y$, then $gcd(x,y)$ divides $m$. Similarly, $gcd(x,y)$ divides $n$ and hence $gcd(x,y)$ divides $gcd(m,n)$.

    Now consider $a_2m - a_1n$ and $b_2m - b_1n$ and conclude that $gcd(m,n)$ divides $x$ and $y$ and hence $gcd(x,y)$.

     – sku May 06 '17 at 22:44

1 Answers1

2

Let $\text{GCD}(x,y) = g$, then we can write $x = gu$, $y = gv$ where $\text{GCD}(u,v) = 1$. \begin{align} m & = a_1x+b_1y = (a_1u+b_1v)g \\ n & = a_2x+b_2y = (a_2u+b_2v)g \end{align} Let $k = \text{GCD}(a_1u+b_1v, a_2u+b_2v)$, we have \begin{align} a_1u+b_1v & = c_1k \\ a_2u+b_2v & = c_2k \end{align} Solve for $u, v$, we get \begin{align} u & = (c_1b_2-c_2b_1)k \\ v & = (c_2a_1-c_1a_2)k \end{align} Since $\text{GCD}(u,v) = 1$, which leads to $k = 1$, or $\text{GCD}(a_1u+b_1v, a_2u+b_2v) = 1$, and $$\text{GCD}(m,n) = g = \text{GCD}(x,y)$$

Guangliang
  • 964