MHB About a variant of the Chinese Remainder Theorem

AI Thread Summary
The discussion revolves around the Chinese Remainder Theorem, specifically addressing two key points. First, it establishes that a system of congruences has a solution if the remainders are congruent modulo the greatest common divisor of the moduli. Second, it seeks to prove that any two solutions to the system are congruent modulo the least common multiple of the moduli. The conversation includes a detailed exploration of the mathematical proof for part (ii), involving the relationships between the integers and their prime factorizations. The notation for the multiplicative inverse in modular arithmetic is clarified, emphasizing its relevance in the context of the theorem.
steenis
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Let $m$ and $m'$ be positive integers, and $d=gcm(m,m')$.
(i) The system:

$x \equiv b (mod \ m)$
$x \equiv b' (mod \ m')$

has a solution if and only if $b \equiv b' (mod \ d)$

(ii) two solutions of the system are congruent $mod \ l$, where $l = lcm(m,m')$.

I can prove part (i), but can anyone help me with part (ii) ?

Remember $gcd(a,b) \cdot lcm(a,b) = a \cdot b$

See, for instance: "Cuoco - Learning Modern Algebra (2013)", p.145 and p.148

Example:
$x \equiv 1 (mod \ 6)$ and $x \equiv 4 (mod \ 15)$
Then $m=6$, $m'=15$, $d=3$, $b=1$, $b'=4$, and $1 \equiv 4 (mod \ 3)$, so (i) applies:
$19 \equiv 1 (mod \ 6)$ and $19 \equiv 4 (mod \ 15)$.
$lcm(6,15)=30$ and all the solutions are: $\cdots \ -41,-11,19,49,79 \ \cdots$.
 
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Hi, steenis.

Start with two solutions, say $x$ and $y$, to the system. We want to show that $l$ divides $x-y$. Since $x$ and $y$ solve the system of congruences, we have that $m$ and $m'$ divide $x-y$, which can be seen by writing $x-y = (x-b)-(y-b)$. Since $m$ and $m'$ both divide $x-y$, so too must their least common multiple, $l$, which can be proved by considering the prime factorizations of $m$ and $m'$.
 
steenis said:
Let $m$ and $m'$ be positive integers, and $d=gcm(m,m')$.
(i) The system:

$x \equiv b (mod \ m)$
$x \equiv b' (mod \ m')$

has a solution if and only if $b \equiv b' (mod \ d)$

(ii) two solutions of the system are congruent $mod \ l$, where $l = lcm(m,m')$.

I can prove part (i), but can anyone help me with part (ii) ?

Hi steenis,

Suppose $\gcd(m,m')=d$, then we can write $m=qd$ and $m'=q'd$ with $\gcd(q,q')=1$.
And suppose $b\equiv b' \pmod d$, so that $\frac{b'-b}d$ is an integer.

Then we have:
$$x = b + km = b' + k'm' \quad\Rightarrow\quad
b + kqd = b' + k'q'd \quad\Rightarrow\quad
kq = \frac{b'-b}d + k'q' \\ \quad\Rightarrow\quad
kq \equiv \frac{b'-b}d \pmod {q'} \quad\Rightarrow\quad
k \equiv [q]_{q'}^{-1} \frac{b'-b}d \pmod {q'} \quad\Rightarrow\quad
k = [q]_{q'}^{-1} \frac{b'-b}d + k''q' \\ \quad\Rightarrow\quad
x= b + km= b + \left([q]_{q'}^{-1} \frac{b'-b}d + k''q'\right)qd = b + [q]_{q'}^{-1}q (b'-b) + k''qq'd
$$

Thus:
$$x \equiv b + [q]_{q'}^{-1}q (b'-b) \pmod{\text{lcm}(m,m')}$$

Btw, note that in the second step we have $b + kqd = b' + k'q'd \quad\Rightarrow\quad (b-b') = (k'q' - kq)d$.
Since the right hand side is divisible by $d$, so must the left hand side be divisible by $d$, proving (i).
 
Thank you, both, very helpful.
However, what does $[q]_{q'}^{-1}$ mean?
 
steenis said:
Thank you, both, very helpful.
However, what does $[q]_{q'}^{-1}$ mean?

The multiplicative inverse of q with respect to q'.
That is, it is such that:
$$[q]_{q'}^{-1} \cdot q \equiv 1 \pmod {q'}$$
And since $q$ and $q'$ are co-prime that inverse is guaranteed to exist and to be unique.

As an example $[2]_5^{-1} = [3]_5$ because $3\cdot 2 \equiv 1 \pmod 5$.

It's a common notation to express solutions that make use of the Chinese Remainder Theorem.
Note that the general solution of $x\equiv a_i \pmod{m_i}$ with $m_i$ pairwise coprime, is $x\equiv\sum a_i \left[\frac M{m_i}\right]_{m_i}^{-1} \frac M{m_i} \pmod M$, where $M=\prod m_i$.
 
Clear, thank you.
I think my texts are too introductory to come across this notation.
 
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