Using Chinese Remainder Theorem when moduli are not pairwise relatively prime

Question

I have this a bit of nightmarish set of equations: $$x \equiv 10\pmod {19271}$$ $$x \equiv 4\pmod {7343}$$ $$x \equiv 8\pmod {9973}$$ For which I am supposed to use Chinese Remainder Theorem to find a solution or to show that one does not exist. Now I have started solving it in a way that would give me $X$ through equation: $$X = r_{1}M_{1}x_{1} + r_{2}M_{2}x_{2} + r_{3}M_{3}x_{3}$$ but then have noticed that $19271$ and $7343$ share a common divisor in 7. The literature I have on my hands doesn't explain that well how to tackle this case, other than having a single remark which states that the theorem used states nothing in the case of moduli not being pairwise relatively prime, and the stuff I went through via online search doesn't really explain it that well due to lack of practical examples.

I was wondering if someone knows how to go about this thing and if it would be possible to explain the steps in a relatively simple way as I had trouble understanding what I encountered so far.

Apply the linked CRT solvability criterion using $,\gcd(19271,7343) = 7\ \ $ — Bill Dubuque, Sep 03 '20 at 18:23

score 3 · Accepted Answer · answered Sep 03 '20 at 07:20

3

In general, if $x=a\pmod{mn}$ then $mn|(x-a)$, so $m|(x-a)$. This means that if $x=a\pmod{mn}$ then $x=a\pmod{m}$.

So since $x=10\pmod{19271}$ then $x=10=3\pmod{7}$. Similarly $x=4\pmod{7343}=4\pmod{7}$. Hence there are no solutions.

answered Sep 03 '20 at 07:20

Chrystomath

10,798

Ah, I see. It seems really simple when it is put that way. Thank you very much! – Rebronja Sep 03 '20 at 08:23

Using Chinese Remainder Theorem when moduli are not pairwise relatively prime

1 Answers1