Suppose $\gcd(a,b)=\gcd(c,d)=\gcd(a,c)=1$. Then, $\{an+b:n\in\mathbb{N}\}\cap\{cn+d:n\in\mathbb{N}\}\neq\emptyset$. Why is that? I think I missed something silly...
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Apply the linked CRT solvability Theorem to the system $, x \equiv b\pmod{a},\ x\equiv d\pmod{c}\ \ $ – Bill Dubuque Dec 16 '20 at 02:23
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Assume $a> 0$ and $c> 0$. Since $\gcd(a,c) = 1$, a multiple of Bézout's identity is
$$ax - cy = d-b$$
for some $x,y\in \mathbb Z$.
To obtain positive $x',y'$ based on $x,y$, note that
$$\begin{align*} ac - ca &= 0\\ a(x+kc) - c(y+ka) &= d-b\\ a(x+kc) + b &= c(y+ka)+d \end{align*}$$
So regardless of whether $x,y\in\mathbb N$, for sufficiently large $k$, both $x'=x+kc \in \mathbb N$ and $y'=y+ka \in \mathbb N$.
peterwhy
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