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We know that if the integers $a$ and $b$ are coprime, then there exist integers $x$ and $y$ such that $xa+yb=1$.

I am wondering whether we could have the following statement true: Statement A: if gcd($a,b$)=$\pm 1$, then there exists an integer $x$ or $y$ (okay if either one of them exists) such that $|xa+b|$ or $|a+yb|$ is a prime number or $1$.

Or could we have the following statement:

Statement B: if gcd($a,b$)=$\pm1$ and both $a$ and $b$ divide $c\neq0$, then there exists an integer $x$ or $y$ such that gcd($xa+b,c$)=$\pm1$ or gcd($a+yb,c$)=$\pm1$. (Again that either one of $x$ and $y$ exists is OK.)

Smart Yao
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    This (your initial question) is essentially Dirchlet’s Theorem on primes in arithmetic progression. (Both $x$ and $y$ exist—because you’re asking for primes $\equiv a\pmod b$ and primes $\equiv b\pmod a$.) See here https://en.m.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions – Jack LeGrüß Nov 13 '20 at 07:31
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    The answer to your second question is always yes when $c=\pm ab$, this holds for all integers $x,y$ coprime to $a,b$. Here, because $a$ and $b$ are coprime both dividing $c$, if a prime $p$ divides both $c$ and $ax+b$ (respectively, $a+by$), then it must necessarily divide either $ax$ or $b$ (respectively, $a$ or $by$). – Jack LeGrüß Nov 13 '20 at 07:54
  • @JackLeGrüß Sorry for my ignorance and your explanation is really helpful! – Smart Yao Nov 13 '20 at 08:03
  • @JackLeGrüß I guess the second statement should be also true by Dirichlet’s theorem you mentioned. We can just make $a+yb$ large enough by a sufficiently large $y$ and Dirichlet’s theorem guarantees the existence of $y$ such that $|a+yb|>|c|$ and $|a+yb|$ is a prime number. – Smart Yao Nov 13 '20 at 08:11
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    When $c\ne\pm ab$, then by the coprimality of $a,b$, it follows that $\frac{c}{ab}$ is an integer other than $\pm 1$. So simply solve for $x$ or $y$ satisfying the congruence $$ax+b\equiv 1\pmod{\frac{c}{ab}},,~,~,~,~,~,~a+by\equiv 1\pmod{\frac{c}{ab}},.$$ – Jack LeGrüß Nov 13 '20 at 08:15
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    Yes, you can use Dirichlet’s Theorem for the second as well; however, that theorem is a nontrivial result to prove, whereas the second question has allow for easy deductions, as expressed in my comments above. – Jack LeGrüß Nov 13 '20 at 08:18
  • @JackLeGrüß Thanks for the supplemental explanation and I will try to digest it. – Smart Yao Nov 13 '20 at 08:22
  • See Dirischlet's theorem for (A) and the linked dupe for (B). – Bill Dubuque Nov 13 '20 at 08:51

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