Let $a,b,c$ be integers such that $(a,b) = 1$, $c>0$. Prove that there is an integer $x$ such that $(a+bx,c) = 1$.
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Have you seen Bezout's identity? – nbritten Nov 28 '19 at 16:03
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Let $p_1,p_2,\ldots$ be the prime factors of $c$. Take $x$ such that $$x\equiv\begin{cases} -\frac{a}{b} & p_i\nmid b \\ 0 & p_i\mid b\end{cases}\pmod{p_i}.$$ This is possible by the Chinese Remainder Theorem. It’s easy to see that this $x$ will satisfy the required condition.
Alternatively, take $x$ such that $a+bx$ is prime: this is possible by Dirichlet’s Theorem.
ViHdzP
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@BillDubuque I mean, our arguments are pretty similar. The $x$ I’m constructing and the $z$ you constructed satisfy the same essential property. – ViHdzP Nov 28 '19 at 18:13
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But there is not use of CRT there (and I show how it follows from a very simple idea) – Bill Dubuque Nov 28 '19 at 18:18