elementary number theory - $aequiv bpmod {m_1}, aequiv bpmod {m_2} implies aequiv bpmod L$

If $\exists a, b, m_1, m_2 \in \mathbb{Z}, a\equiv b\pmod {m_1}$ and $a\equiv b\pmod {m_2}$, then $\exists k \in \mathbb{Z}, a=b+{m_1m_2k\over gcd(m_1,m_2)} \implies a\equiv b\pmod L$, where $L$ is the l.c.m. of $m_1$ and $m_2$.

I am trying to derive, starting with the first part, with no success

1. $\exists l \in \mathbb{Z}, a -b = l\cdot m_1 \implies a =b +l\cdot m_1$
   


2. $\exists m \in \mathbb{Z}, a-b = m\cdot m_2 \implies a =b +m\cdot m_2$
   

Equating both, $b +l\cdot m_1 = b +m\cdot m_2$

Unable to start the proof

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Based on comment by @saulspatz, the new attempt to proof is:
$\exists r_1, r_2 \in \mathbb{Z},$ s.t.
(i) $m_1 = r_1\gcd(m_1, m_2),$ (ii) $m_2 = r_2\gcd(m_1, m_2)$
Multiplying, (i) by (ii), we get:
$m_1m_2 = r_1r_2(m_1, m_2)^2$
if for suitable integer $k$, take $r_1r_2={1\over k}$, then not possible as $r_1, r_2$ are themselves integers.

So, anyway attempt something:
Let, $r_1r_2=k'$, $m_1m_2 = k'(m_1, m_2)^2 \implies (m_1, m_2) = {m_1m_2 \over {k'\cdot (m_1, m_2)}}$

But, $k'$ is an integer, and the final form has an integer $k$ in numerator.

Answer

Let $M = m*n$ be a multiple of $n$.

Then $a \equiv b \mod n \implies a \equiv b + kn \mod M$ for some $k: 0 \le k

So if we have $L = \text{lcm}(m_1,m_2) = k_1m_1 = k_2m_2$ then

Then if $a \equiv b \mod m_1$ and $a\equiv b \mod m_2$ then $a \equiv b + l_2*m_1 \mod L$ and $a \equiv b + l_2*m_1 \mod L$ where $l_1 < k_1$ and $l_2 < k_2$.

So $l_1*m_1 \equiv l_2*m_2$. Now $0\le l_1*m_1 < k_1m_1 = L$ and $0\le l_2*m_2 , k_2m_2 = L$ so $l_1*m_1 = l_2*m_2$ so $l_1m_1 = l_2m_2$ is a common multiple of $m_1, m_2$.

But $L$ is the least common multiple so $l_1*m_1 = l_2*m_2 = 0$.

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$a \equiv b \mod m_1$ means $a = b + j*m_1 = b+ j*\gcd(m_1,m_2)*\frac {m_1}{\gcd(m_1,m_2)}$.

And $a\equiv b \mod m_2$ means $a = b + l*m_2 = b + l*\gcd(m_1,m_2)*\frac {m_2}{\gcd(m_1,m_2)}$

So $j*\gcd(m_1,m_2)*\frac {m_1}{\gcd(m_1,m_2)}=l*\gcd(m_1,m_2)*\frac {m_2}{\gcd(m_1,m_2)}$

So $j*\frac {m_1}{\gcd(m_1,m_2)}=l*\frac {m_2}{\gcd(m_1,m_2)}$

But $\frac {m_1}{\gcd(m_1,m_2)}$ and $\frac {m_2}{\gcd(m_1,m_2)}$ are relatively prime.

So $\frac {m_1}{\gcd(m_1,m_2)}|l$ and $\frac {m_2}{\gcd(m_1,m_2)}|j$

So $j*\frac {m_1}{\gcd(m_1,m_2)}=l*\frac {m_2}{\gcd(m_1,m_2)}= k*\frac {m_1}{\gcd(m_1,m_2)}*\frac {m_2}{\gcd(m_1,m_2)}$

And $j*\gcd(m_1,m_2)*\frac {m_1}{\gcd(m_1,m_2)}=l*\gcd(m_1,m_2)*\frac {m_2}{\gcd(m_1,m_2)}= k*\gcd(m_1,m_2)*\frac {m_1}{\gcd(m_1,m_2)}*\frac {m_2}{\gcd(m_1,m_2)}=k*\frac {m_1m_2}{\gcd(m_1,m_2)}$

So $a = b+ k*\frac {m_1m_2}{\gcd(m_1,m_2)}$

And $a \equiv b \mod \frac {m_1m_2}{\gcd(m_1,m_2)}$

And $L = \frac {m_1m_2}{\gcd(m_1,m_2)} =$ lowest common multiple of $m_1, m_2$.