A linear combination of `a` and `b` is some integer of the form , where .

There's a very interesting theorem that gives a useful connection between linear combinations and the GCD of `a` and `b`, called Bézout's identity:

**Bézout's identity:** (the GCD of `a` and `b`) is the smallest positive linear combination of non-zero `a` and `b`.

Both Bézout's identity and its corollary I show below are very useful tools in elementary number theory, being used for the proofs of many of the most fundamental theorems. Let's see why it's true.

**(I) Intuition:** First I'd like to explain this (surprising at first sight) theorem intuitively. By defintion, any common divisor of `a` and `b` will divide for all . In particular, also divides any .

Now, assume we've found some small which isn't the GCD. But we've just said that divides all linear combinations, so it also divides `x`. Therefore, `x` can not be smaller than the GCD. In other words, the smallest positive linear combination can only be itself.

**Corollary:** An integer is a linear combination of `a` and `b` IFF it is a multiple of their GCD.

To prove Bézout's identity more formally, and along the way to see why the corollary is also true, let's first prove the following:

**(III)** Let `I` be a nonempty set of integers that is closed under addition and subtraction, and contains at least one non-zero integer. Then there exists a smallest positive element , and `I` consists of all multiples of `b` ().

Proof: `I` contains at least one non-zero integer. Then it definitely contains at least one positive integer, because it is closed under addition and subtraction. Assume we have for some . Therefore and then also . Thus we have positive integers in `I`. According to the well-ordering principle, `I` has a smallest positive element which we'll call `b`.

Now we'll want to show that . As usual, to prove equalities of sets, it will be shown that they contain one another.

is obvious - since `I` contains `b` and is closed under addition and subtraction, it contains all the multiples of `b`.

To prove we'll demonstrate that any element is a multiple of `b`. Using the division algorithm we write for some integers `q` and `0 <= r < b`. But this means that (because `I` contains `bq` and `c` and is closed under subtraction and addition). However, recall that `b` was chosen to be the smallest positive element of `I`, so `r` must be equal to 0. Therefore `c` is a multiple of `b`, and we have shown that . *Q.E.D.*

Now back to Bézout's identity. We'll define:

This `I` is obviously non-empty and is closed under addition and subtraction (by its definition as a linear combination). Note, in particular, that it also contains `a` and `b`. By **(III)**, `I` consists of all multiples of its smallest positive element, which we'll call `d` here.

To show that we have to show that `d|a`, `d|b` and if `c|a` and `c|b` then `c|d`. First, by definition `d` is a divisor of any element in `I`, so it also divides `a` and `b`. If `c|a` and `c|b`, say `a=cq` and `b=cp`, then:

So `d|c`, which completes our proof that `d=(a,b)`. *Q.E.D.*

Regarding the corollary, it stems trivially from the definition of `I` and the proof above.