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I believe that the following statement is true, but I couldn't prove it.

Let $u,m,k\in \Bbb Z$. If $\gcd(u,m)=1$ then there exists an integer $k_1\in\Bbb Z$ such that $\gcd(mk_1+u,mk)=1$ Thanks.

Bernard
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Ergin Süer
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    Why do you believe this? – miracle173 Jul 13 '18 at 08:17
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    We have a natural surjective ring homomorphism $\Bbb Z/mk\Bbb Z\to \Bbb Z/m\Bbb Z$ given by $a+mk\Bbb Z\mapsto a+m\Bbb Z$. And we know that the restriction map of $(\Bbb Z/mk\Bbb Z)^{\times}$ into $(\Bbb Z/m\Bbb Z)^{\times}$ is also surjective. – Ergin Süer Jul 13 '18 at 08:29

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You assertion holds for $k\neq 0$, for by Dirichlet's theorem the arithmetic progression $u+hm$ for $h\in\Bbb Z$ contains infinitely many prime numbers hence at least one of them doesn't divide (and, thus, it is prime with) $mk$.

Alternatively, take $k_1$ to be the product of prime divisors of $k $ which doesn't divide $u $.

See also here.

Fabio Lucchini
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