整数整除性证明求助：若d|2a+b且d|3a+b，求证d|a且d|b

Hey there! Let's break this down step by step—you're already halfway there by starting with the right assumptions. The key here isn't to solve for a and b directly, but to use a core property of divisibility that makes this proof straightforward.

First, let's recall a critical divisibility rule: If d | x and d | y, then for any integers m and n, d | (mx + ny). In plain terms, if d divides two numbers, it divides any integer linear combination of those numbers.

Step 1: Prove d | a #

We know:

• d | (2a + b) (so there's some integer k₁ where 2a + b = d*k₁)
• d | (3a + b) (so there's some integer k₂ where 3a + b = d*k₂)

Let's subtract the first equation from the second to eliminate b:

(3a + b) - (2a + b) = d*k₂ - d*k₁

Simplify both sides:

• Left side: 3a + b - 2a - b = a
• Right side: d*(k₂ - k₁)

Since k₂ - k₁ is an integer (integers are closed under subtraction), this means a = d*(k₂ - k₁). By definition of divisibility, this tells us d | a.

Step 2: Prove d | b #

Now that we know d | a, we can use this with one of our original divisibility statements. Let's take d | (2a + b).

Since d | a, we also know d | 2a (if d divides a number, it divides any integer multiple of that number). Using the divisibility rule again: if d | (2a + b) and d | 2a, then d | [(2a + b) - 2a].

Simplify the expression inside:

(2a + b) - 2a = b

So this means d | b.

Alternative Shortcut for d | b #

If you want to skip using the d | a result first, you can use another linear combination:

3*(2a + b) - 2*(3a + b) = 6a + 3b - 6a - 2b = b

Since d | (2a + b) and d | (3a + b), d divides the left-hand side of this equation, so it must divide the right-hand side (which is b). That directly gives d | b, and you can then use similar logic to prove d | a.

Either way, we've shown that if d | 2a + b and d | 3a + b, then d | a and d | b.

内容的提问来源于stack exchange，提问作者DavidM