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4.4 Applications of Bézout’s Theorem

In this section we shall explore some of the theoretical consequences of Bézout’s Theorem.

Our first proposition combines several statements from Section 4.2 into a single result.

Proposition 4.4.1

Let a,b∈ℤ∖{0}a,b\in\mathbb{Z}\setminus\{0\}, and let d∈ℕd\in\mathbb{N} be a common factor of aa and bb. Then the following three conditions are equivalent:

  1. (a)

    dd is the highest common factor of aa and b;b;

  2. (b)

    dd is an integral linear combination of aa and b;b;

  3. (c)

    every common factor of aa and bb is a factor of dd.

Proof. Theorem 4.2.17 shows that (a) implies (b), while Lemma 4.2.13 ensures that (b) implies (c), and finally (c) implies (a) by Lemma 4.2.8. □\Box

We shall next give an explicit description of all the integral linear combinations of a pair of non-zero integers.

Proposition 4.4.2

Let a,b∈ℤ∖{0}a,b\in\mathbb{Z}\setminus\{0\}. Then the integral linear combinations of aa and bb are precisely the multiples of hcf⁢(a,b)\mathrm{hcf}(a,b).

Example 4.4.3

If we take a=18a=18 and b=30b=30, we know that hcf⁢(a,b)=6=2⁢a-b;\mathrm{hcf}(a,b)=6=2a-b; since for instance 24=4⋅624=4\cdot 6 we have 24=4⁢(2⁢a-b)=8⁢a-4⁢b=8⋅18-4⋅30.24=4(2a-b)=8a-4b=8\cdot 18-4\cdot 30.

Proof. Let d=hcf⁢(a,b)d=\mathrm{hcf}(a,b) and n∈ℤn\in\mathbb{Z}. We can then rewrite the statement as follows:

(∃r,s∈ℤ)(n=ra+sb)⇔(d|n).(\exists\,r,\,s\in\mathbb{Z})(n=ra+sb)\ \Leftrightarrow\ (d|n).

‘‘⇒\Rightarrow’’. Suppose that n=r⁢a+s⁢bn=ra+sb for some r, s∈ℤr,\,s\in\mathbb{Z}. Then, as d|ad|a and d|b,d|b, Lemma 4.2.13 implies that d|nd|n.

‘‘⇐\Leftarrow’’. Suppose that d|nd|n, say n=m⁢dn=md for some m∈ℤm\in\mathbb{Z}. By Bézout’s Theorem, we can write d=u⁢a+v⁢bd=ua+vb for some u, v∈ℤu,\,v\in\mathbb{Z}, and therefore we have

n=m⁢d=(m⁢u)⁢a+(m⁢v)⁢b=r⁢a+s⁢bn=md=(mu)a+(mv)b=ra+sb

provided that we define r=m⁢ur=mu and s=m⁢vs=mv. □\Box

Proposition 4.4.4

Let a,b∈ℤ∖{0}a,b\in\mathbb{Z}\setminus\{0\} and c∈ℕc\in\mathbb{N}. Then hcf⁢(c⁢a,c⁢b)=c⋅hcf⁢(a,b)\mathrm{hcf}(ca,cb)=c\cdot\mathrm{hcf}(a,b).

Example 4.4.5

Since hcf⁢(12,20)=4\mathrm{hcf}(12,20)=4, we have hcf⁢(120,200)=10⋅4=40.\mathrm{hcf}(120,200)=10\cdot 4=40.

Proof. Let d=hcf⁢(a,b)d=\mathrm{hcf}(a,b); then d|ad|a and d|bd|b so cd|cacd|ca and cd|cbcd|cb, so that c⁢dcd is a common factor of c⁢aca and c⁢bcb. By Theorem 4.2.17 d=r⁢a+s⁢bd=ra+sb for some r,s∈ℤr,s\in\mathbb{Z}, so c⁢d=r⁢(c⁢a)+s⁢(c⁢b)cd=r(ca)+s(cb); thus Corollary 4.2.15 implies that hcf⁢(c⁢a,c⁢b)=c⁢d.\mathrm{hcf}(ca,cb)=cd. □\Box

Corollary 4.4.6

Let a,b∈ℤ∖{0}a,b\in\mathbb{Z}\setminus\{0\} with d=hcf⁢(a,b)d=\mathrm{hcf}(a,b), and write a=d⁢αa=d\alpha and b=d⁢βb=d\beta, where α,β∈ℤ\alpha,\beta\in\mathbb{Z}. Then hcf⁢(α,β)=1\mathrm{hcf}(\alpha,\beta)=1.

Example 4.4.7

hcf⁢(27,45)=9\mathrm{hcf}(27,45)=9; we have 27=9⋅327=9\cdot 3 and 45=9⋅545=9\cdot 5, and 33 and 55 are coprime.

Proof. By Proposition 4.4.4 we have

d=hcf⁢(a,b)=hcf⁢(d⁢α,d⁢β)=d⋅hcf⁢(α,β),d=\mathrm{hcf}(a,b)=\mathrm{hcf}(d\alpha,d\beta)=d\cdot\mathrm{hcf}(\alpha,% \beta),

so as d≠0d\neq 0, hcf⁢(α,β)=1\mathrm{hcf}(\alpha,\beta)=1 as required. □\Box

Our final result is called Euclid’s Lemma; it will be important in our study of prime numbers in Section 4.6.

Theorem 4.4.8

Let a,b,p∈ℤ∖{0}a,b,p\in\mathbb{Z}\setminus\{0\}, and suppose that p|abp|ab and that aa and pp are coprime. Then p|bp|b.

Example 4.4.9

60=5⋅1260=5\cdot 12; as 6|606|60 and hcf⁢(6,5)=1\mathrm{hcf}(6,5)=1, we have 6|12.6|12.

Proof. Since p|abp|ab, we can take q∈ℤq\in\mathbb{Z} with p⁢q=a⁢bpq=ab. As hcf⁢(p,a)=1\mathrm{hcf}(p,a)=1, Theorem 4.2.17 implies that there are r,s∈ℤr,s\in\mathbb{Z} such that r⁢p+s⁢a=1rp+sa=1. Then

b\displaystyle b =b⋅1=b(rp+sa)=brp+sab\displaystyle=b\cdot 1=b(rp+sa)=brp+sab
=brp+spq=p(br+sq),\displaystyle=brp+spq=p(br+sq),

and so p|bp|b as required. □\Box