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Consider the following system of modular equations: $${x}\bmod{6}=4$$ $${x}\bmod{7}=5$$ $${x}\bmod{8}=6$$ (Problem 2.16 c) from Gorodentsev's Algebra textbook). If the moduli were coprime, the system would be solvable by the standard use of the Chinese remainder theorem. However, 6 and 8 are not coprime, so the CRT is not applicable. Does that necessarily imply that the system has no solutions? If it doesn't, is there a general method of finding those? Note that this question doesn't answer mine.

Bill Dubuque
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Daigaku no Baku
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    @JohnOmielan, yes, you are right, sorry, fixed – Daigaku no Baku Jun 30 '25 at 22:46
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    No worries. Regarding your "Does that necessarily imply that the system has no solutions?", the answer is no. With your particular example, you can easily confirm that $x=168k-2$, for any integer $k$, are all solutions (the $168$ comes from $168=\operatorname{lcm}(6,7,8)$. Note that they are also actually the only solutions. – John Omielan Jun 30 '25 at 22:50
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    ${x}\bmod{6}=4$ is equivalent to ${x}\bmod{3}=1$ and ${x}\bmod{2}=0$ both being true, though ${x}\bmod{8}=6$ also implies ${x}\bmod{2}=0$, so you could solve ${x}\bmod{3}=1$ and ${x}\bmod{7}=5$ and ${x}\bmod{8}=6$ in the usual way – Henry Jun 30 '25 at 22:57
  • Please search for answers before asking questions. – Bill Dubuque Jun 30 '25 at 23:26

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One strategy you can use is to split each modular equation into its prime power factors. For instance, $x \mod 6 = 4$ becomes $x \mod 2 = 0$, $x \mod 3 = 1$.

Then for each $p$, it's easy to check if the equalities modulo powers of $p$ are consistent - for instance, if something is $6 \mod 8$ then it is $0 \mod 2$ also.

If the system is consistent, then regular CRT on the largest power of each prime (in this case, $1 \mod 3$, $5 \mod 7$, $6 \mod 8$) gives you the family of solutions.

mathperson314
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