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Let $a$, $b$, and $c$ be integers with $\gcd(a, b, c) = 1$. Prove that there is some integer $x$ such that $\gcd(a + xb, c) = 1$.

I know I should use the Chinese remainder theorem, but I'm still a little stuck on this proof.

Lisa
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user191465
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  • Let $x = c$... Can you detail the question? – Tacet Nov 10 '14 at 23:44
  • I think that $x = c$ does not work because any prime dividing $a$ and $c$ divides both $c$ and $a+cb$. As an example consider $a = 2\cdot 3,\ b = 3 \cdot 5,\ c = 5 \cdot 2$. – jflipp Nov 11 '14 at 08:59

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This can be proved using Dirichlet's theorem on primes in arithmetic progressions (http://en.wikipedia.org/wiki/Dirichlet's_theorem_on_arithmetic_progressions), although that theorem is not so elementary ...

The proof goes as follows. Put $d := gcd(a,b),\ a' := a/d,\ b' := b/d$. Then $gcd(a',b') = 1$, so Dirichlet's theorem applies, and the sequence $a'+kb'$ for $k \in \mathbb N$ contains infinitely many primes. So we can find a $k'$ such that $gcd(a'+k'b',c) = 1$, for example we can pick $k'$ such that $a'+k'b'$ is prime and larger than $c$. Moreover, $gcd(a,b,c) = 1$ implies $gcd(d,c) = 1$. All in all, we get $gcd(d(a'+k'b'),c) = 1$ and $d(a'+k'b') = a+k'b$. So $k'$ is the desired integer.

jflipp
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