2

If we know $c,d\in\mathbb Z/N\mathbb Z$ and $\gcd(c,d,N)=1$. How do we prove there always exist $k_1, k_2\in\mathbb Z$ such that $\gcd(c+k_1N, d+k_2N)=1$ ?

It looks simple but I've been stuck for long. Thanks.

Bill Dubuque
  • 272,048
JKDASF
  • 547
  • Hint: Bezout's identity is an "if and only if" – Mummy the turkey Dec 05 '20 at 08:40
  • @Mummytheturkey Hi. I tried to use Bezout's identity, that is, to prove $\gcd(+_1,+_2)=1$, I try to prove there is a $\mathbb Z$ linear combination of these two numbers that equals to $1$. Use the assumption, I know there is always a linear combination of $c,d,N$ that equals to $1$. But then I am stuck as I cannot always split the term with $N$ to add them as part of $c$ and $d$. Can you elaborate a little bit more on that? Thank you – JKDASF Dec 05 '20 at 17:12
  • anything unclear – reuns Dec 06 '20 at 22:01
  • By a simple proof: $(c,N,d)=1\Rightarrow \exists k!: (\overbrace{c!+!k N}^{\large C},d)=1.,$ Applied again $\exists j!: (C,d!+!jN)=1,$ by $(C,d,N) = (c!+!k N,d,N)=(c,d,N)=1.\ \ $ – Bill Dubuque Dec 10 '20 at 11:10

1 Answers1

1

Let $$s = \prod_{p\ |\ c,p\ \nmid\ N} p$$

Take $u$ such that $$uN=1-d\bmod s$$

$d+uN=d\bmod \gcd(c,N)$ is a unit and $d+uN=1\bmod s$ is a unit thus $$\gcd(c,d+uN)=1$$

reuns
  • 77,999