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Given two integers $x$ and $y$,

Is $\gcd(x,y)=1$ equivalent to $ p \not \mid x $ for EVERY prime divisor of $y$

the actual question is to find all integers $k$ such that

$$\gcd(k+8,18)=1$$

thanks.

RobPratt
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ahmed
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    Your equivalence is correct (consider what it means for $\gcd(x,y) \gt 1$), but this might be challenging to apply to solve "the actual question". Good luck! – hardmath Mar 20 '20 at 14:53
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    this way $18 = 2*3^2 $ $k$ is odd and $k+8\not\equiv 0 \pmod 3$ – ahmed Mar 20 '20 at 14:55
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    Yes, if a list of integers is coprime to $x$ then so too is their product (by the inductive extension of the linked dupe). Yours is the special case when it is a list of primes (factors of $y).,$ The converse is clear (if $y$ is coprime to $x$ then so too is every factor of $y)\ \ \ $ – Bill Dubuque Mar 20 '20 at 15:06
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    Note that you could also prove the result directly using FTA (existence and uniqueness of prime factorizations), but you'll soon need the general result so you may as well become familiar with that since the proof is easy (using Bezout or gcd laws as in the linked answers). – Bill Dubuque Mar 20 '20 at 15:15

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Notice that if $k$ satisfies the condition, that is, $\gcd(k+8,18)=1,$ then since $\gcd((k+18)+8,18)=1,$ $k+18$ also satisfies the condition.

Thus, you only need to consider solutions to this equation for $0\le k<18,$ and every other solution is a multiple of $18$ plus one of these "base" solutions.

Kenta S
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    Note that the equivalence observed by the OP allows us to simply work mod $6$ rather than mod $18$. – hardmath Mar 20 '20 at 15:01