求证：整除c-a与c+a的正整数为1或2及相关GCD等式

开发者阿泽

2026-5-19

Proof for the Number Theory Problem

Let's break this down clearly — first, we need to establish our context: we're working with primitive positive integer Pythagorean triples (i.e., (a, b, c) are positive integers satisfying (c^2 = a^2 + b^2) and (\text{GCD}(a, b, c) = 1)). For such triples, (c) is always odd, and exactly one of (a) or (b) is odd (since the sum of two odd squares would be congruent to 2 mod 4, which can't be a perfect square).

Part 1: Any positive integer dividing both (c-a) and (c+a) must be 1 or 2

Let (d) be a positive integer that divides both (c-a) and (c+a). By basic divisibility rules, (d) must divide any linear combination of these two numbers — specifically their sum and difference:

(d \mid (c+a) + (c-a) = 2c)
(d \mid (c+a) - (c-a) = 2a)

This means (d) divides (\text{GCD}(2a, 2c) = 2 \cdot \text{GCD}(a, c)). Now, since we're dealing with a primitive triple:

If (\text{GCD}(a, c) = g > 1), then (g) would divide (c^2 - a^2 = b^2), so (g \mid b) — which contradicts (\text{GCD}(a, b, c) = 1). Thus, (\text{GCD}(a, c) = 1).

So (\text{GCD}(2a, 2c) = 2), which means (d) can only be a positive divisor of 2. The only such divisors are 1 and 2.

Part 2: Prove (\text{GCD}\left(\frac{c-a}{2}, \frac{c+a}{2}\right) = 1)

First, since (c) and (a) are both odd (from the primitive triple property), (c-a) and (c+a) are both even — so (\frac{c-a}{2}) and (\frac{c+a}{2}) are integers.

Let (g = \text{GCD}\left(\frac{c-a}{2}, \frac{c+a}{2}\right)). Again, using divisibility rules for sums and differences: