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Let's say we have $a\equiv x\mod b$ and $a\equiv y\mod c$. Is there any way to find the relationship between $x$ and $y$ for nontrivial (a>b and/or a>c) moduli?

How about a specific example, such as

$a\equiv x\mod b$ vs $a\equiv y\mod 2b$,

or

$a\equiv x\mod b$ vs $a\equiv y\mod 2b-1$.

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    $\begingroup$ Note that $x$ and $y$ are not uniquely defined as integers, but are representatives of equivalence classes (as you have stated the problem) - perhaps you want the least non-negative residue? $\endgroup$ – Mark Bennet Oct 2 '18 at 19:52
  • $\begingroup$ Sorry for the flippant description. The relationship between the equivalence classes is indeed what I am going for (I mean to say that determining any element in the equivalence class will do). Least non-negative residue would be great if you could do so. $\endgroup$ – Tejas Rao Oct 2 '18 at 19:57
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For any integer $k \neq -1, 0, 1$ $$a \equiv y \pmod {(k b)} \qquad \textrm{implies} \qquad a \equiv y \pmod b,$$ but the converse does not hold in general. On the other hand $$a \equiv y \pmod b \qquad \textrm{implies} \qquad a \equiv y + \ell b \pmod {(kb)}$$ for some integer $\ell$ (in fact, a unique integer in $\{0, \ldots, k - 1\}$).

On the other hand, $b$ and $k b - 1$ are coprime for any integers $k, b$ such that $b \neq 0$, $(k, b) \neq (\pm 1, \pm 1)$, so the Chinese Remainder Theorem tells us that for any remainders $m$ and $n$ we can find an integer $a$ such that $$a \equiv m \pmod b \qquad \textrm{and} \qquad a \equiv n \pmod {(2 b - 1)} .$$ In this sense there is no relation between residues modulo $b$ and $k b - 1$. Conversely, however, given coprime $p, q$ we can use the residue classes modulo those numbers to determine the residue class modulo $pq$.

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