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Given two systems of congruences with one common parameter $A$ I need to find all values of $A$ for which both systems have the same solution sets. The systems are: $$x\equiv5\bmod6\\x\equiv A\bmod35$$and$$x\equiv A\bmod10\\x\equiv 14\bmod21$$

I solved it by first finding x in both systems, by CRT and then since the answers were both $\bmod 210$ I equated both solutions with that modulus and solved for $A$.
From the first system I got $x\equiv36A+35\bmod210$, and from the second one $x\equiv21A-70\bmod210$. Equating the $x$'s I got $A\equiv3\bmod6$. Is it correct? And what would I have to do if the moduli weren't equal? Would finding the LCA and multiplying both solutions be correct?

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  • $\begingroup$ The solutions of the systems are correct $\endgroup$ – Peter Aug 30 '17 at 12:42
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    $\begingroup$ For $A=3$, the expressions $36A+35$ and $21A-70$ are not congruent modulo $210$ . $A\equiv 7\mod 14$ seems to be correct. $\endgroup$ – Peter Aug 30 '17 at 12:48
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    $\begingroup$ @Peter Thanks, I double-checked it again and now I got your answer. $\endgroup$ – Joald Aug 30 '17 at 12:52
  • $\begingroup$ The case where the moduli are not equal : If they are coprime, you can apply CRT again, have to think about the remaining case $\endgroup$ – Peter Aug 30 '17 at 12:53
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There is an error somewhere. It is simpler to split into prime moduli using CRT:

$\qquad \begin{align} x\equiv\, \color{#c00}1\!\!\!\pmod 2,\ &\ 2\!\!\!\pmod 3,\ A\!\!\!\pmod 5,\ \color{#0a0}A\!\!\!\pmod 7\\ x\equiv \color{#c00}A\!\!\!\pmod 2,\ &\ 2\!\!\!\pmod 3,\ A\!\!\!\pmod 5,\ \ \color{#0a0}0\!\!\!\pmod 7 \end{align}$

By CRT these are equivalent $\iff \color{#c00}A\equiv \color{#c00}1\pmod 2,\ \color{#0a0}A\equiv \color{#0a0}0\pmod 7\,$

Solving we find $\bmod 2\!:\,\ 1\equiv A\equiv 7k\equiv k\,$ so $\, A = 7k = 7(1\!+\!2n) = 7\!+\!14n$

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  • $\begingroup$ Does this method always find every possible solution, and if so, why? $\endgroup$ – Joald Aug 30 '17 at 13:00
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    $\begingroup$ @Joald Yes, because by CRT the split systems are equivalent to the originals, and the two completely split systems are equivalent iff they have the same prime moduli and same remainders for each prime, again by CRT. Think about that carefully to be sure you understand how it follows from CRT. $\endgroup$ – Bill Dubuque Aug 30 '17 at 13:05

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