数论推导疑问：由a≡b(mod m₁)、a≡b(mod m₂)推导a≡b(mod L)

Hey, let's work through this step by step to get past that stuck point you hit! First, let's recap your initial progress, then build on it:

由∃l∈ℤ，a−b = l·m₁得a = b + l·m₁；由∃m∈ℤ，a−b = m·m₂得a = b + m·m₂。将两式相等得b + l·m₁ = b + m·m₂，至此无法继续推导...

Great start—you've correctly translated the congruences into integer equations, and now we just need to leverage properties of the greatest common divisor (gcd) to move forward. Here's how to do it:

1. Simplify the equality
   First, subtract b from both sides of your final equation to get:
   l·m₁ = m·m₂

2. Introduce the gcd of m₁ and m₂
   Let d = gcd(m₁, m₂). By definition, we can write:

   • m₁ = d·m₁' where m₁' = m₁/d
   • m₂ = d·m₂' where m₂' = m₂/d
     Crucially, gcd(m₁', m₂') = 1 (we've factored out all common divisors).

3. Substitute back into the equation
   Replace m₁ and m₂ in l·m₁ = m·m₂:
   l·d·m₁' = m·d·m₂'
   We can cancel out d from both sides (since d is a positive integer, this is valid):
   l·m₁' = m·m₂'

4. Use coprimality to find divisibility
   Since gcd(m₁', m₂') = 1, m₂' must divide l (this follows from Euclid's lemma, generalized to coprime integers: if a coprime pair divides a product, each divides one factor). So there exists some integer k ∈ ℤ such that:
   l = k·m₂'

5. Substitute back to find a - b
   Recall that a - b = l·m₁. Plug in l = k·m₂' and m₁ = d·m₁':
   a - b = k·m₂'·d·m₁'
   Rearrange the terms:
   a - b = k·d·m₁'·m₂'

6. Connect to the least common multiple (lcm)
   Remember that the lcm of m₁ and m₂ is given by:
   lcm(m₁, m₂) = (m₁·m₂)/gcd(m₁, m₂) = (d·m₁'·d·m₂')/d = d·m₁'·m₂'
   Let L = lcm(m₁, m₂). Then our equation becomes:
   a - b = k·L
   Which directly translates to the congruence:
   a ≡ b (mod L)

And if we want to write a explicitly in terms of b, m₁, m₂, and k:
a = b + k·(m₁·m₂)/gcd(m₁, m₂)
Which matches the result you mentioned at the start!

内容的提问来源于stack exchange，提问作者jitender