# Lecture 34: Properties of GCD

As we saw last lecture, the set of units of $$ℤ_m$$ is quite important. In this lecture we will show how to compute the greatest common divisor and how it relates to units.

## Greatest common divisor

Let $$g : ℕ \times ℕ → ℕ$$ be defined inductively on its second input as follows: $$g(a,0) := a$$ and $$g(a,b) = g(b,r)$$ where $$r$$ is the remainder of $$a$$ divided by $$b$$. Note that this inductive definition is reasonable in the same way that a proof by strong induction is reasonable, because $$r \lt b$$; you might say this is a "strongly inductively" defined function.

I claim that $$g(a,b)$$ is the greatest common divisor of $$a$$ and $$b$$. We will prove this inductively. Note that the form of induction here follows the form of induction we used to define $$g$$; in this case $$g(a,b)$$ is defined by strong induction on $$b$$, and our proofs use strong induction on $$b$$.

The proof proceeds in two parts: First, it is a common divisor; Second, it is greater than any other common divisor.

Claim 1: $$g(a,b)$$ divides $$a$$ and $$g(a,b)$$ divides $$b$$.

Proof: By strong induction. Let $$P(b)$$ be the statement "for all $$a$$, $$g(a,b)$$ divides both $$a$$ and $$b$$".

$$P(0)$$ is clear: $$g(a,0) = a$$; since $$a = a \cdot 1$$, $$g(a,0)$$ divides $$a$$; while $$0 = a \cdot 0$$ so $$g(a,0)$$ divides $$0$$.

We now prove $$P(b)$$ assuming $$P(k)$$ for all $$k \lt b$$. For brevity, let $$g = g(a,b)$$. We want to show that $$g \mid a$$ and $$g \mid b$$. Well, by definition, $$g = g(b,r)$$ where $$r$$ is the remainder of $$a$$ by $$b$$ (in other words, $$a = qb + r$$ and $$0 \leq r \lt b$$). Since $$r \lt b$$, we have assumed $$P(r)$$, so we know that $$g(b,r) \mid b$$ and $$g(b,r) \mid r$$. This immediately shows us that $$g = g(b,r) \mid b$$, so all that's left to show is that $$g \mid a$$.

Since $$g(b,r)$$ divides $$b$$ and $$r$$, we have $$b = kg$$ and $$r = ℓg$$. We also know $$a = qb+r = qkg + ℓg = (qk + ℓ)g$$, which shows $$g \mid a$$ as required.

Now we show that $$g$$ is the greatest common divisor:

Claim 2: $$g(a,b)$$ is the greater than any other common divisor of $$a$$ and $$b$$. In other words, if $$c \mid a$$ and $$c \mid b$$ then $$g(a,b) \geq c$$.

We will actually prove something stronger:

Claim 2': if $$c \mid a$$ and $$c \mid b$$ then $$c \mid g(a,b)$$.

This is stronger because if $$a \mid b$$ then $$b \geq a$$.

Proof of 2': Again, by strong induction on $$b$$. Choose an arbitrary $$c$$. Let $$P(b)$$ be the statement "for all $$a$$, if $$c \mid a$$ and $$c \mid b$$ then $$c \mid g(a,b)$$".

To see $$P(0)$$, assume that $$c \mid a$$ and $$c \mid 0$$. Well, $$g(a,0) = a$$, so $$c \mid g(a,0)$$ by assumption.

Now, assume $$P(k)$$ for all $$k \lt b$$, and assume that $$c \mid a$$ and $$c \mid b$$. We wish to show that $$c \mid g(a,b)$$. As above, we'll write $$g = g(a,b)$$ for brevity. Now, $$g(a,b) = g(b,r)$$ where $$a = qb+r$$. Since we are interested in $$g(b,r)$$, we want to use our inductive hypothesis $$P(r)$$. $$P(r)$$ says "for any $$a$$, $$c \mid g(a,r)$$", so we will choose $$a = b$$, so that we have "if $$c \mid b$$ and $$c \mid r$$ then $$c \mid g(b,r)$$".

In order to use this fact, we must show that $$c \mid r$$ (we already know $$c \mid b$$). Well, since $$c \mid a$$ and $$c \mid b$$, we have $$a = kc$$ and $$b = ℓc$$ for some $$k$$ and $$ℓ$$. Since $$a = qb + r$$ we have $$r = a - qb = kc - qℓc = (k - qℓ)c$$, so $$c \mid r$$. Therefore, $$P(r)$$ gives us $$c \mid g(b,r) = g(a,b)$$ as required.

## Bézout coefficients

The following result is helpful for relating the gcd to the set of units:

Claim 3 (Bézout's identity): For all $$a$$ and $$b \in ℤ$$, there exist constants $$s$$ and $$t$$ such that $$gcd(a,b) = sa + tb$$.

$$s$$ and $$t$$ are referred to as the Bézout coefficients of $$a$$ and $$b$$.

Before the proof, let us see how this relates to units.

Corollary: $$[a]_m$$ is a unit if $$gcd(a,m) = 1$$. Indeed, since $$1 = sa + tm$$, we can take equivalence classes mod $$m$$ on both sides we get $$$1$ = $s$$a$ + $t$$m$ = $s$$a$$$ since $$[m]_m = [0]$$. Thus $$[s]$$ is the inverse of $$[a]$$.

Proof of claim: By strong induction on $$b$$. Let $$P(b)$$ be the statement "for all $$a$$, there exists $$s$$ and $$t \in ℤ$$ such that $$gcd(a,b) = sa + tb$$.

$$P(0)$$ is obvious, because $$gcd(a,0) = a = 1 \cdot a + 0 \cdot 0$$; thus we can choose $$s = 1$$ and $$t = 0$$.

To see $$P(b)$$, assume $$P(k)$$ for all $$k \lt b$$. Then we have $$gcd(a,b) = gcd(b,r) = s'b + t'r$$ for some $$s'$$ and $$t'$$ (by $$P(r)$$). Since $$a = qb + r$$, we have that $$r = a - qb$$. Plugging this in, we see $$gcd(a,b) = s'b + t'(a - qb) = t'a + (s' - t'q)b$$. Therefore, choosing $$s = t'$$ and $$t = s' - t'q$$ gives the result.