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I have an easy question of arithmetic.

Let $a, b, N$ be integer numbers such that $\mathrm{gcd}(a,b,N) = 1$. Is it true that there exists an integer number $x \in \mathbb{Z}$ such that $\mathrm{gcd}(a+Nx,b) = 1$?

Although the question seems easy, I cannot find a proof. Is anyone helping me?

Andrea
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  • $b$ isn't needed. If $(a,N)=1$, then $(a,N+a)=1$. – T.J. Gaffney Mar 05 '14 at 09:37
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    @Gaffney: $(a,b,N)=1$ does not imply $(a,N)=1$. – jpvee Mar 05 '14 at 09:39
  • This doesn't look easy at all. Hm. Could be very interesting. – Guy Mar 05 '14 at 09:44
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    Asked before at least twice: http://math.stackexchange.com/questions/609810/show-that-if-a-b-and-c-are-integers-such-that-a-b-1-then-there-i and http://math.stackexchange.com/questions/647600/prove-if-gcda-b-c-1-then-there-exists-z-such-that-gcdazb-c-1. The first link isn't exactly the same question but the proofs given there can be used here too. – Bart Michels Mar 05 '14 at 10:52

1 Answers1

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This is a classical consequence of the Chinese remainder theorem. Denote by $p_1,p_2,\ldots,p_r$ the prime divisors of $b$.

For each $i$, $p_i$ does not divide both $a$ and $N$.

I claim that there is a $x_i\in{\mathbb Z}$ such that $a+Nx_i$ is not divisible by $p_i$.

Indeed, if $p_i$ divides $N$ then it does not divide $a$ by the above so any $x_i$ will do. If, on the other hand, $p_i$ does not divide $N$, then $N$ is invertible modulo $p_i$ and we have an $x_i$ such that $a+Nx_i\equiv 1 \ ({\sf mod} \ p_i)$.

By the Chinese remainder theorem, there is an $x\in{\mathbb Z}$ such that $x\equiv x_i \ ({\sf mod} \ p_i)$ for every $i$. This $x$ will do.

Ewan Delanoy
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