What to do if the modulus is not coprime in the Chinese remainder theorem?

Question

Chinese remainder theorem dictates that there is a unique solution if the congruence have coprime modulus.

However, what if they are not coprime, and you can't simplify further?

E.g. If I have to solve the following 5 congruence equations

$x=1 \pmod 2$

$x=1 \pmod 3$

$x=1 \pmod 4$

$x=1 \pmod 5$

$x=1\pmod 6$

as gcd (2,3,4,5,6) is not coprime, how would you do it. I have heard that you can't use the lcm of the numbers, but how does it work?

Sorry for the relatively trivial question and thank you in advance.

Answer 1

Break into coprime modulo-s

$$x\equiv1\pmod6\implies x\equiv1\pmod2, x\equiv1\pmod3\ \ \ \ (1)$$

$$x\equiv1\pmod2,x\equiv1\pmod4\implies x\equiv1\pmod4\ \ \ \ (2)$$

$$x\equiv1\pmod5\ \ \ \ (3)$$

Now apply CRT on $1,2,3$

But as all the residues are same, we don't need CRT.

We only need $x-1$ to be divisible by lcm$(2,3,4,5,6)$

Answer 2

$\newcommand{\lcm}{\mathrm{lcm}}$The general method is the following. Suppose you have a system of two congruences $$\tag{two} \begin{cases} x \equiv a \pmod{m}\\ x \equiv b \pmod{n}\\ \end{cases} $$ This is soluble iff $\gcd(m, n) \mid a - b$. If this is the case, first find, using Euclid's algorithm, $u, v$ such that $$ m u + n v = \gcd(m, n). $$ Multiply by $$ \lambda = \frac{a - b}{\gcd(m, n)} $$ to get $$ m (u \lambda) + n (\lambda v) = a - b, $$ and now a solution is $$ x = a - m (u \lambda) = b + n (\lambda v) = \sigma, $$ and all solutions are the numbers $$\tag{one} x \equiv \sigma \pmod{\lcm(m, n)} $$ If you have more than two equations, use this method on the first two to reduce (two) to (one), so you have an equation less. Repeat.

Answer 3

We need to eliminate the pairs that are not coprime to ensure we don't get a contradiction (otherwise no such solution can exist).

To check this, we must ensure that if $x \equiv a \mod{m}$ and $x \equiv b \mod{n}$, then $a \equiv b \mod{\gcd(m,n)}$.

This means that asking that $x \equiv 1 \mod{2}$ and $x \equiv 1 \mod{4}$ is fine since $1 \equiv 1 \mod{2}$, whereas if we required $x \equiv 2 \mod{4}$ instead, then this is not true meaning no such solution can exist.

If these conditions are all compatible then we can get a solution modulo the lowest common multiple of all these numbers.

To do this, note that by Chinese Remainder Theorem (CRT), if $x \equiv a \mod{mn}$, where $\gcd(m,n)=1$, then this is equivalent to requiring that $x \equiv a \mod{m}$ and $x \equiv a \mod{n}$.

Repeating this, we can assume $m$, $n$ are prime powers. If we haven't checked compatibility yet, we can do at this point as it will be quite straightforward. Assuming they are, if we have a congruence condition modulo $p^m$, we can eliminate any other congruence condition modulo $p^n$ for any $n<m$.

Now all our moduli will be coprime and we can now apply CRT again to get a solution (although we may have more equations than we started with).

In your example, splitting each congruence into prime powers we have:

\begin{eqnarray} &x& \equiv 1 \mod{2}, \\ &x& \equiv 1 \mod{3}, \\ &x& \equiv 1 \mod{4}, \\ &x& \equiv 1 \mod{5}, \\ &x& \equiv 1 \mod{2} \mbox{ and } x \equiv 1 \mod{3}. \end{eqnarray}

Now we may remove the last line since these conditions already appear. The only other check we need to make is the congruences for $2$ and $4$. Now these are compatible so we remove the first condition. This leaves us with the conditions

\begin{eqnarray} &x& \equiv 1 \mod{3}, \\ &x& \equiv 1 \mod{4}, \\ &x& \equiv 1 \mod{5}. \\ \end{eqnarray}

Now all the moduli are coprime so by CRT we find the solution is $x \equiv 1 \mod{60}$.

Answer 4

\begin{align} x &\equiv 1 \pmod 2 \\ x &\equiv 1 \pmod 3 \\ x &\equiv 1 \pmod 4 \\ x &\equiv 1 \pmod 5 \\ x &\equiv 1 \pmod 6 \\ \end{align}

Note that $x \equiv 1 \pmod 6 \implies \left\{ \begin{array}{l} x\equiv 1 \pmod 2 \\ x \equiv 1 \pmod 3 \end{array} \right.$

Replace $x \equiv 1 \pmod 6$ in your original list with those two equivalences and sort the list by prime number bases:

\begin{align} x &\equiv 1 \pmod 2 \\ x &\equiv 1 \pmod 2 \\ x &\equiv 1 \pmod 4 \\\hline x &\equiv 1 \pmod 3 \\ x &\equiv 1 \pmod 3 \\ \hline x &\equiv 1 \pmod 5 \\ \end{align}

Note first that $x \equiv 1 \pmod 4 \implies x \equiv 1 \pmod 2$. This means that the equivalence $x \equiv 1 \pmod 2$ is included in the equivalence $x \equiv 1 \pmod 4$ and is therefore superfluous - it can be removed.

So we can simplifiy the list to

\begin{align} x &\equiv 1 \pmod 4 \\ x &\equiv 1 \pmod 3 \\ x &\equiv 1 \pmod 5 \\ \end{align}

The solution is $x \equiv 1 \pmod{30}$.