0

On my textbook, after the proof of the Chinese Remainder Theorem when the moduli is pairwise relatively prime, one modification is left as exercise.

Prove that the system of linear congruences $x\equiv a_i\pmod{m_i}$ is solvable if and only if $gcd(m_i,m_j) \vert (a_i - a_j)$ for all $i\neq j$, and $i,j \in \{ 1,2,\dots r \} .$ (or $a_i\equiv a_j\mod{gcd(m_i,m_j)}$ which is the same). If a solution exists, it is unique $\mod{lcm(m_1,m_2,...,m_r)}$

I managed to find a proof for the case $r=2$, and from there tried to do induction, but I am stuck when I have to use the inductive hypothesis.

$[ \longrightarrow ]$

Suppose that $a_i\equiv a_j\mod{(m_i,m_j)}$ for all $i\neq j$, and $i,j \in \{ 1,2,\dots,r, r+1 \}$.

Then, the system

$$ x\equiv a_1 \mod{m_1} \\ x\equiv a_2 \mod{m_2}\\ .\\.\\.\\ x\equiv a_r \mod{m_r}\\ $$

has a solutions, lets say $b$ (by inductive hypohesis).

I will next like to consider the system

$$ x\equiv b \mod{lcm(m_1,m_2,...,m_3)}\\ x\equiv a_{r+1} \mod{m_{r+1}} $$

and find that it has a solution using the case $r=2$, but I don't know how to use my hyphotesis here because I need $b \equiv m_{r+1}\mod{gcd(lcm(m_1,m_2,...,m_r),m_{r+1}))}$

$[\longleftarrow]$: As for the converse I don't know how to begin.

Bill Dubuque
  • 272,048
Camilo Diaz
  • 33

1 Answers1

2

Since $x\equiv b$ is a solution to the first system, $b\equiv a_i \ mod \ m_i$.
Note that $a_i \equiv a_{r+1} \ mod \ gcd(m_i,m_{r+1})$ and $gcd(m_i,m_{r+1})$ divides $m_i$, we get: $$b\equiv a_i \equiv a_{r+1} \ mod \ gcd(m_i,m_{r+1})$$ for all $1 \leq i \leq r$.
Then, $$b \equiv a_{r+1}\mod \ lcm(gcd(m_1,m_{r+1}),...,gcd(m_r,m_{r+1}))=A $$ Now it remains to show that $A$ is divisible by $B=gcd(lcm(m_1,...,m_r),m_{r+1})$
Let $p$ be any prime number, we'll show that $v_p(A) \geq v_p(B)$, where $v_p(x)$ is the multiplicity of $p$ in the prime factorization of $x$.
We can check that $v_p(A)$ is the largest number in $min(v_p(m_i),v_p(m_{r+1}))$ where $1 \leq i \leq r$
and $v_p(B)$ is the smaller number between $max(v_p(m_1),...,v_p(m_r))$ and $v_p(m_{r+1})$
There're only two cases:
$\textbf{Case 1}$: $v_p(m_{r+1}) \gt v_p(m_i)$ for all $1 \leq i \leq r$
Then $v_p(A)$ the largest number in $v_p(m_1),...,v_p(m_r)$ i.e $v_p(A)=max(v_p(m_1),...,v_p(m_r))$
$v_p(m_{r+1}) \gt max(v_p(m_1),...,v_p(m_r)) $ so $v_p(B)=max(v_p(m_1),...,v_p(m_r)) $
$\textbf{Case 2}$: $\exists \ 1 \leq k \leq r: v_p(m_{r+1}) \leq v_p(m_k)$
Then $v_p(A) \geq v_p(m_{r+1}) $ because $min(v_p(m_k),v_p(m_{r+1}))=v_p(m_{r+1})$
$v_p(B)=v_p(m_{r+1})$ because $v_p(m_{r+1}) \leq v_p(m_k) \leq max(v_p(m_1),...,v_p(m_r)) $
In both cases, $v_p(A) \geq v_p(B)$ so $B$ divides $A$ and we're done.
The converse is actually much easier, assume that $x \equiv b \ mod \ lcm(m_1,...,m_r)$ is a solution.
Then $b \equiv a_i \ mod \ m_i$, $ b \equiv a_j \ mod \ m_j$ for all $1 \leq i \lt j \leq r$
But $gcd(m_i,m_j)$ divides both $m_i$ and $m_j$ so $gcd(m_i,m_j)$ divides $(a_i-b)+(b-a_j)=a_i-a_j$

Jenn Inn
  • 54