Let $n$ be an integer. If $(a,m)=1$, there exists an integer $a'$ such that $a'\equiv a \pmod{m}$ and $(a',n)=1$.

Question

Let $n$ be an integer. If $(a,m)=1$, there exists an integer $a'$ such that $a'\equiv a \pmod{m}$ and $(a',n)=1$.

I am not sure if the above statement is true. It has been annoyingly elusive to both proof and counterexample. Intuitively, it seems like it should be true, and there are no small counterexamples (I have done some computer analysis). An example would be taking $n=12,m=4$ and $a=3$. In this case $a'=7$ satisfies the required properties. Of course, any such $a'$ needs to be of the form $mk+a$, so the problem seems to boil down to choosing $k$ with sufficient cleverness. At this point I've thrown the kitchen sink at it to no avail, so any ideas/solutions would be greatly appreciated!

Answer 1

We show that we can find infinitely many $k$ such that $mk+a$ is relatively prime to $n$. And we do not use the condition $m\mid n$.

Let $P$ be the product of all the primes that divide $n$ but do not divide $m$ (this could be an empty product, in which case $P=1$.)

Since $\gcd(m,P)=1$, the congruence $mx+a\equiv 1\pmod{P}$ has a solution $s$, and $s+tP$ is a solution for any integer $t$. Let $k=s+tP$.

We show that $mk+a$ is relatively prime to $n$ by showing that if the prime $p$ divides $n$, then $p$ does not divide $mk+a$.

Suppose first that $p$ divides $m$. Then $p$ cannot divide $mk+a$, for $m$ and $a$ are relatively prime.

Now suppose $p$ does not divide $m$. Then $p$ divides $P$, and therefore $ms+a\equiv 1\pmod{p}$. Thus $mk+a\equiv ms+a\equiv 1\pmod{p}$, so again $p$ does not divide $mk+a$. It follows that $mk+a$ and $n$ are relatively prime. Since there are no restrictions on $t$, there are infinitely many numbers of the form $mk+a$ that are relatively prime to $n$.

Answer 2

Let $\,n = jm,\ a' = a\!+\!km.\,$ Then $(a',n) = (a\!+\!km,jm) = (a+km,j)\,$ by $\,(a\!+\!km,m) = \color{#c00}{(a,m)=1}.\ $ By $ $ Stieltjes, $\,(\color{#c00}{a,m},j)=1\,\Rightarrow\,(a\!+\!km,j)= 1\,$ for some $k$.

Remark $\ $ See also this challenge question on whether this can be proved without using prime factorizations. It is currently an open problem if this can be done using only Bezout's identity for the gcd (i.e in Bezout domains). It is known for special cases (e.g. one-dimensional), but the problem has been open since Kaplansky introduced closely related properties when studying rings which admit Smith normal forms.