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Is it possible to apply Chinese Remainder Theorem to the follow system of congruences?
$$\begin{align}x&\equiv1\mod 15\\ x&\equiv2\mod 21\\ x&\equiv3 \mod 35\end{align}$$ $15, 21$ aren't coprime
$21$ and $35$ aren't coprime
Is CRT applicable to this problem?

Awnon Bhowmik
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Juan
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2 Answers2

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$$x \equiv 1 \pmod{15} \iff \begin{array}{c} x \equiv 1 \pmod 3 \\ x \equiv 1 \pmod 5 \end{array} $$

$$x \equiv 2 \pmod{21} \iff \begin{array}{c} x \equiv 2 \pmod 3 \\ x \equiv 2 \pmod 7 \end{array} $$

$$x \equiv 3 \pmod{35} \iff \begin{array}{c} x \equiv 3 \pmod 5 \\ x \equiv 3 \pmod 7 \end{array} $$

Note that there are contradictions with the values of $x \pmod 3, x \pmod 5, \ \text{and} \ x \pmod 7$.

Hence there can be no solution.

Steven Alexis Gregory
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The system doesn't have a solution, You can apply CRT to $x \equiv 1 \bmod 15$ to get $x \equiv 1 \bmod 3$ and $x \equiv 1 \bmod 5$, and since $5$ and $21$ are coprime, you can combine $x \equiv 1 \bmod 5$ and $x \equiv 2 \bmod 21$ to get $x \equiv 86 \bmod 105$, which is incompatible with $x \equiv 3 \bmod 35$.

Connor Harris
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    Isnt it simpler ? $3$ divides $15$ and $21$ so $1\equiv 2(\mod 3)$. – Rene Schipperus May 07 '18 at 01:18
  • Is there a quick way to see if a system does not have a solution? Do I just check to see if the modulus are coprime or not? – Juan May 07 '18 at 01:18
  • @ReneSchipperus that is a faster method, you're right – Connor Harris May 07 '18 at 01:18
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    @Juan systems with non-coprime moduli can have solutions, but they don't have to. For example, $x \equiv 2 \bmod 4, x \equiv 4 \bmod 6$ has a solution, but not (say) $x \equiv 2 \bmod 4, x \equiv 3 \bmod 6$. The simple method is to apply CRT to break every congruence into a system of congruences modulo powers of primes, and then check whether the resulting system is inconsistent. – Connor Harris May 07 '18 at 01:20
  • What exactly do you mean by inconsistent? – Juan May 07 '18 at 01:22
  • For an example, in the system that you posed, $x \equiv 1 \bmod 15$ implies $x \equiv 1 \bmod 3$ and $x \equiv 2 \bmod 21$ implies $x \equiv 2 \bmod 3$, and this is inconsistent (i.e., doesn't have a solution). – Connor Harris May 07 '18 at 01:23