Proof that if a natural number is coprime to two other natural numbers than there is another coprime natural number within a given range. Let $n$, $m$, and $c$ be distinct natural numbers such that $1 < n < m$ and $c$ is coprime to both $n$ and $m$.
Show that there is a natural number $d$ coprime to both $n$ and $m$ such that $c < d \leq c+2m$.
I can’t quite put my finger on how to figure out how to prove it.
I am also curious if it generalizes to any number of natural numbers all coprime to $c$, but I feel like that might be a bit challenging for MSE. So a proof of a generalization is not necessary if it is known to be beyond reasonable effort of an answerer.
 A: We prove a stronger claim. First, we may assume WLOG that $\gcd(m,n)=1$. This is because you can remove any prime factors dividing $m$ from the prime factorization of $n$ and the claim is unaffected, since only $n$ decreases, but the union of the prime factors is still the same.
Next, we show that given positive integer $k$, the positive integers $km+1,km+2,\ldots,(k+1)m$ contain a number that is relatively prime to $m$ and $n$. Write $n=ab$ where $a$ is the product of all primes (with multiplicity) dividing $n$ that divide $k$, and $b$ is the product of all primes dividing $n$ that do not divide $k$. Consider the number $km+b$:

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*For any prime $p \mid m$, we have $p \mid km$. Since $p \nmid n$, we have $p \nmid b$ and hence, $p \nmid (km+b)$.

*For any prime $p \mid a$, we have $p \mid k$, and hence $p \mid km$. Since $p \nmid b$, we have $p \nmid (km+b)$.

*For any prime $p \mid b$, we have $p \mid n$, and hence $p \nmid m$. Moreover, as $p \mid b$, we have $p \nmid k$. Thus, $p \nmid (km+b)$.

In conclusion, we can see that none of the primes dividing $m$ or $n$ divide  $km+b$, which is clearly between $km+1$ and $(k+1)m$ since $km < km+b \leqslant km+n \leqslant km+m$. This solves the problem, since for any choice of $c$, the interval $(c,c+2m]$ contains positive integers $km+1,km+2,\ldots,(k+1)m$ for some positive integer $k$.
