In the existence proof of the standard answer to the number theory problem , how are the solutions constructed? It seems rather elusive.

Question

This number theory problem is from my textbook and we can’t use Dirichlet theorem .[If $k>0,l>0,gcd(k,l)=1$,then there are infinite primes in the sequence $\left \{ kn+l\right \} $]

problem:If $a,b\in \mathbb{Z} ,b\ne 0,gcd(a,b)=1$,show that for a $m \in \mathbb{Z^{*}}$,there are infinite solutions for the equation about $k$, $gcd(a+bk,m)=1,k \in \mathbb{Z}$

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standard answer (there is only the proof for existence, and in the follow ,the proof of infinite is easy):
We suppose $c$ is the largest number which satisfies $gcd(c,a)=1$ and $c |m$.
If $gcd(a+bc,m)=d$,we will prove $gcd(a+bc,m)=1$.
As $gcd(a,b)=1,gcd(a,c)=1$,we get $gcd(a,bc)=1$. So we have $gcd(d,a)=1,gcd(d,bc)=1$

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If not,then $gcd(d,a)>1$ or $gcd(d,bc)>1$
Then,$gcd(d,a)$ or $gcd(d,bc)$ have prime divisor，i.e.,exist prime $p$,
s.t. $p|gcd(d,a)$ or $p|gcd(d,bc)$
As $d|a+bc$,
If $p|gcd(d,a)$，then $p|a,p|d$,so $p|a$ and $p|bc$,it contradicts with $gcd(a,bc)=1$.
If $p|gcd(d,bc)$, similarly,it contradicts with $gcd(a,bc)=1$.
So $gcd(d,bc)=1,gcd(d,c)=1$.

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In the other hand,as $d|m,c|m,gcd(d,c)=1$,
we have $dc|m$.
Also because $gcd(d,a)=gcd(a,c)=1$,
we have $gcd(a,cd)=1$.
As $c$ is the largest number which satisfies $gcd(c,a)=1$ and $c |m$,the number $d$ must be $0$(If not,$dc>c$).
In conclusion, we have completed the proof of existence.
confusion
How to find the number $c$ which is defined in beginning of this answer ?

Answer 1

It might be easier to envision what is going on if you look at the prime factorizations.

If $a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ is the prime factorization of $a,$ and $m=p_1^{m_1}\cdots p_k^{m_k}p_{k+1}^{m_{k+1}}\cdots p_n^{m_nx},$ where some of $m_1,m_2,\dots,m_k$ might be zero, then $c$ is defined to strip out all the primes that divide $a.$ Then, the definition of $c$ gives $$c=p_{k+1}^{m_{k+1}}\cdots p_n^{m_n}$$

Then if prime $p\mid c,$ then $p$ is not a factor of $a,$ so is not a factor of $a+bc.$

And if prime $p\mid m/c,$ then $p\mid a$ and $p\not\mid bc,$ so $p$ is not a factor of $a+bc.$

But if prime $p\mid m,$ then $p\mid c$ or $p\mid m/c,$ and thus, in either event, $p\not\mid a+bc.$

So $\gcd(m,a+bc)=1.$

Answer 2

The task is impossible if $m = 0$, so you need to specify that $m \not= 0$. It is also impossible if $b = 0$ (which implies $a = \pm 1$), in the sense that there are not infinitely many numbers $a + bk$ as $k$ varies, which I think is the intention in the problem, even though there are infinitely many $k$. So we should assume $b \not= 0$.

Taking these corrections into account, you want to show that if $a, b, m$ are integers such that $b$ and $m$ are nonzero and $\gcd(a,b) = 1$, then there is an integer $a' \equiv a \bmod b$ such that $(m,a') = 1$. Once we have one such $a'$, the infinitely many numbers $a' + bmx$ as $x$ runs over $\mathbf Z$ are all congruent to $a \bmod b$ and are all relatively prime to $m$. When $b = \pm 1$ we can use $a' = 1$, so we may assume $|b| \geq 2$.

In the second paragraph of the proof of Theorem 3.2 here, I use the Chinese remainder theorem (not Dirichlet's theorem) to show that if $a, b$, and $N$ are integers such that $a \not= 0$, $|N| \geq 2$, and $\gcd(a,b,N) = 1$, then there is an integer $b' \equiv b \bmod N$ such that $\gcd(a,b') = 1$.

Let's change the notation in the previous paragraph to match yours: rewrite the previous paragraph's $a$ as $m$, $b$ as $a$, and $N$ as $b$. Then that argument shows that if $m$, $a$, and $b$ are integers such that $m \not= 0$, $|b| \geq 2$, and $\gcd(m,a,b) = 1$, then there is an integer $a' \equiv a \bmod b$ such that $\gcd(m,a') = 1$. In particular, we can find such an $a'$ when $\gcd(a,b) = 1$, which is a stronger condition than $\gcd(m,a,b) = 1$.