# Greatest Common Divisor

The problem is simple: given two positive integers, \(a\) and \(b\), how do we find \(\gcd(a,b)\), the greatest common divisor of \(a\) and \(b\) (i.e. the largest integer that \(a\) and \(b\) are both multiples of)?Just about everybody learns how to do this at school. You break each of \(a\) and \(b\) down into a product of prime factors (where a prime number is one with exactly two factors, itself and \(1\)) and then multiply together all the common prime factors to get the biggest number that each is a multiple of. But this actually relies on quite a deep result. The procedure works because

**The Fundamental Theorem of Arithmetic**says that any positive integer can be expressed as a product of primes, and it can only be done in one way (except for the order of the factors).

So, there are two issues here.

The first is in plain sight. It is the question of why you should believe the fundamental theorem of arithmetic. This might not seem like an issue. In fact, after a lot of experience in factorizing integers you may well feel that it is pretty obvious. But it isn't, and part of the intention of this post is to persuade you that it isn't.

The other is hidden, but comes into view when the numbers \(a\) and \(b\) get larger. It is the problem that working out the prime factorization of a number is really quite hard when the number is large.

There is, fortunately, an answer for both of these:

**Euclid's algorithm**gives us a different way to compute the greatest common divisor of two integers, and, as a bonus, we can use it to prove that every integer has a unique prime factorization. This algorithm is to be found in Euclid's Elements, and is probably the oldest non-trivial algorithm in significant current use, being a vital ingredient in the RSA public key cryptosystem.

## Euclid's Algorithm

The algorithm relies on a familiar fact from basic arithmetic, which goes by the appallingly inappropriate name of**the division algorithm**. It isn't an algorithm: an algorithm is a set of step-by-step instructions which tell you how to do something. This just tells you that a certain type of question has a particular type of answer. A couple of sources call it

**the division theorem**because of this, but, alas, the terminology is very well established. I would rather be right than popular, though, so I will call it the division theorem.

But enough whining. What does this result say?

**The Division Theorem**If \(a\) and \(b\) are positive integers, then there is a unique positive integer \(q\), and a unique non-negative integer \(r \lt b\) such that \(a=qb+r\).

In other words, if you divide \(a\) by \(b\) you get a quotient and remainder, and there's only one right answer. This is something so entrenched in our experience that we take it very much for granted, and that's just what I'm going to do. But it has consequences.

If the remainder, \(r\), is zero, then \(a\) is a multiple of \(b\): we also say that \(b\) divides \(a\).

Back to the original question: I have two positive integers, \(a\) and \(b\), and I want to know the biggest number that divides into both of them. Let's suppose that \(a \gt b\). (If not, I can swap them over). Now, by the division theorem, I know that I can write \(a\) as \(qb+r\), where \(0\le r \lt b\). What's more, if \(d\) is any number that divides \(a\) and \(b\), since \(a-qb=r\), \(d\) divides \(r\). And since \(a=qb+r\), if \(d\) divides into \(b\) and \(r\), it also divides \(a\). That means that \(b\) and \(r\) have exactly the same common divisors as \(a\) and \(b\).

But something really useful has happened. I've replaced the problem of finding the greatest common divisor of \(a\) and \(b\) by finding the greatest common divisor of \(b\) and \(r\), and \(r\) is smaller than \(b\), which is smaller than \(a\). I've made the problem easier.

There are two possibilities: if \(r=0\), then \(b\) divides into \(a\), so it is the greatest common divisor of \(a\) and \(b\). If not, I can repeat the trick. Eventually the remainder will be \(0\), and I will have the greatest common divisor of my original \(a\) and \(b\).

This procedure is the essence of Euclid's algorithm. Let's see how it works with an example.

What is \(\gcd(60,25)\)? We calculate as follows: \[ \begin{split} 60 &= 2 \times 25 + 10\\ 25 &= 2 \times 10 + 5\\ 10 &= 2 \times 5 \end{split} \] This tells us that \(\gcd(60,25)=5\). But we get something not at all obvious from it. We can rewind the calculation like this: \[ \begin{split} 5 &= 25 - 2 \times 10\\ &= 25-2 \times (60-2 \times 25)\\ &= 5 \times 25 - 2 \times 60 \end{split} \] This tells us that we can express \(5\) as a combination of \(60\) and \(25\). But there's nothing special about these numbers: I could have rewound any appliation of Euclid's algorithm in just the same way.

So, given any \(a\) and \(b\), we can use Euclid's algorithm to find \(\gcd(a,b)\), and also to find integers \(m,n\) with the property that \(\gcd(a,b)=ma+nb\). This wasn't

*at all*obvious how to do from the prime factorization approach. In fact, with the prime factorization approach it would never have occurred to me that this could even be done.

Thinking about this approach to finding the gcd of two numbers, we can see that \(\gcd(a,b)\) isn't just the biggest number that divides into both \(a\) and \(b\): it is actually a multiple of any other divisor. The reasoning is just what was used above: any divisor of \(a\) and \(b\) is also a divisor of \(r\), the remainder. Repeating this, we eventually find that it must be a divisor of the last non-zero remainder, i.e. it must be a divisor of \(\gcd(a,b)\).

You might wonder who cares? I think there are two immediate reasons to care, namely that this addresses both of the issues I raised earlier. The first reason is that you don't have to take it on trust that prime factorization works as advertised, since you can see just why this works. The second is more practical. It is that for large numbers, this algorithm is very fast: enormously fast compared to the best known algorithms for factorizing large numbers, which is important. It is the fact that factorizing large integers is (as far as we know) expensive while finding the greatest common divisor is cheap that makes RSA public key cryptography work.

There are other, less immediate reasons too. Using this, we can actually prove that prime factorization works as advertised. But before getting there, we need to see something useful that this tells us about prime numbers.

## Prime numbers

Suppose that \(p\) is a prime number, and \(a\) is any number which is not a multiple of \(p\). Then we must have \(\gcd(a,p)=1\), which tells us that there are integers \(m,n\) such that \(ma+np=1\). Now comes the clever bit.Let \(p\) be a prime number, and suppose that \(a\) and \(b\) are integers such that \(ab\) is a multiple of \(p\). Then either \(a\) or \(b\) must be a multiple of \(p\).

"But that's obvious!" you may be tempted to object. But if you think it's obvious, think about why: it's almost certainly because you're so accustomed to how prime factorization works. Unfortunately that's exactly the thing that I want to prove.

So, let's suppose that \(ab\) is a multiple of \(p\). If \(a\) is a multiple of \(p\), we're done. But what if \(a\) is not a multiple of \(p\)? Well, in that case \(\gcd(a,p)=1\), so there are integers \(m,n\) such that \(ma+np=1\). Now multiply this equation by \(b\). We get \[ b=mab+npb. \] But \(ab\) is a multiple of \(p\) and \(npb\) is a multiple of \(p\), so their sum - which is \(b\) - must also be a multiple of \(p\). That's what we needed.

Now we have everything we need to prove the

# Fundamental Theorem of Arithmetic

## At last, the proof

We can now consider an integer, \(n\). There are two possibilities: either \(n\) is a prime, or \(n\) is the product of two smaller factors. Then we think about each of them in turn. Eventually, we must reach a point where each of our factors is a prime.But this is only part of the problem, and the easy part at that. How do we know that there is only one way to express \(n\) as a product of primes?

This is where we use that last fact. Suppose we have two prime factorizations of \(n\), say \[ n=p_1p_2 \ldots p_k = q_1q_2 \ldots q_l \] Then \(p_1\) is a factor of \(q_1q_2 \ldots q_l\). If \(p_1\) is not a factor of \(q_1\) (in which case \(p_1=q_1\)), then \(p_1\) is a factor of \(q_2\ldots q_l\), and by repeating this argument we find that \(p_1\) must be one of the \(q_i\). Then divide \(n\) by \(p_1\) and repeat the argument. Eventually we find that each of the prime factors in either factorization must also be a factor in the other factorization, so the two are actually the same (except possibly in a different order).

## What is a prime number again?

I rushed past this idea up at the top, with the brief statement that a prime number is one with exactly two factors, itself and \(1\). This isn't quite satisfactory if we allow negative numbers, so let's be a little more general than that, and try to give a definition which will work in more generality.We say that a number is a

**unit**if it has a multiplicative inverse, and then that a factorization is

**proper**if no factor is a unit. Then a prime is a number with no proper factorization. So this works when we allow negative numbers, since although we can write \(3=1\times 3 = -1 \times -3\), in each case one of the factors is a unit.

But this property wasn't really what we used to prove the fundamental theorem of arithmetic. We used this to show that any integer has a prime factorization, but not to show that it was unique. For that part of the argument, the important fact was that if a product is a multiple of a prime, then at least one of the factors in the product must be a multiple of the prime.

In fact there two ways of characterizing prime numbers.

One characterization is that \(p\) is prime if it has no proper factorization.

The other is that \(p\) is prime if, whenever \(ab\) is a multiple of \(p\) then at least one of \(a\) and \(b\) must be a multiple of \(p\). We saw above that every prime has this property: we just have to make sure that any number with this property is prime. But if \(n\) has a proper factorization into \(ab\) then \(n\) divides into \(ab\), but since \(n\) is larger than both \(a\) and \(b\) neither is a multiple of \(n\), so no number with a proper factorization can satisfy this property.

But it doesn't have to be like that.

## Another number system

The integers aren't the only number system we could work with. For example, we might consider the Gaussian integers \(\mathbb{Z}[i]\), i.e. the complex numbers whose real and imaginar parts are both integers. More interestingly, we could, and we will, do something rather more exotic. We consider numbers of the form \(m+n\sqrt{-5}\), where \(m\) and \(n\) are both integers, and call this collection \(\mathbb{Z}[\sqrt{-5}]\). Note that as for the integers, the units here are just \(\pm 1\).We now get something happening which cannot happen with the integers.

First, notice that \(|m+n\sqrt{-5}| = \sqrt{m^2+5n^2}\), and since these objects are complex numbers, if \(M,N \in \mathbb{Z}[\sqrt{-5}]\), \(|MN|=|M||N|\). Then in this number system, just as in \(\mathbb{Z}\), neither \(2\) nor \(3\) has a proper factorization; and neither do \(1 \pm \sqrt{-5}\). You can check this in each by writing down all the numbers with modulus less than the target value, and checking that you can't get the target by multiplying any collection of them together (unless one is a unit). But \[ 2 \times 3 = (1+\sqrt{-5})(1-\sqrt{-5}) = 6. \] So in this number system we see that \(1+\sqrt{-5}\) is a factor of \(2 \times 3\) but it is not a factor or either \(2\) or \(3\).

So in this case we do not get a unique factorization. It can't

*really*be obvious that there is only one way to factorize a number into a product of terms which can't be factorized any further, because there are number systems where it doesn't happen!

## Prime versus irreducible

In this more general context, it is necessary to split up our two notions of prime, because they no longer coincide. The standard way to do this is to call a number with no proper factorization an**irreducible**number, and to define a prime to be a number which has the property that if it is a factor of \(ab\) then it must be a factor of at least one of \(a\) or \(b\). (This is why on your first encounter with abstract algebra, the definition of prime can seem to have nothing to do with the familiar prime numbers.)

Then any prime number must be irreducible, but a number can be irreducible without being prime. For each number to have a unique prime factorization, it must also be the case that every irreducible number is prime.

# Moving on

This may have already been more than enough for you. But if not, there are lots of extenstions to play with here.We could play the game again with the Gaussian integers, and it turns out that again there is a unique factorization into irreducibles. And we could consider variations, such as numbers of the form \(m+n\sqrt{-k}\) where \(k\) is a positive integer. How does the choice of \(k\) affect the behaviour? We already know that when \(k=5\), unique factorization fails. But is this typical, or does it happen for other vaues? Is there a property of \(k\) that we can calculate that tells us?

We could try with polynomials with real coefficients: it turns out that division of polynomials is similar enough to division of integers that we have a division theorem and a Euclidean algorithm, that prime and irreducible coincide, and that any polynomial can be expressed uniquely as a product of primes (which here means linear factors or quadratics with no roots). We could ask how things change if we take different types of coefficients, say complex, or rational, or integer.

These two alone give a large toy box to play with and explore these ideas, but they're only a start. Explore, investigate, make up your own variations. If all else fails, look in a book on abstract algebra or ask the internet.

# Calculating gcd redux

So, after all that, how should you actually calculate the gcd of two integers?The answer is, as it almost always is, "it depends". If the numbers are small enough to be factorized easily, using the prime factorization is probably the quicker of the two. If they are large enough that factorization is a pain, then Euclid's algorithm will do the job with less pain.

And I think it's worth calculating a few both ways just to see how the same result is arrived at by such very different routes.

How come that r−qb=a ?

ReplyDeleteIncompetence, sir or madam, sheer incompetence. Fixed now, and thanks for picking it up.

ReplyDeleteNo problem, thanks for nice article. ;)

ReplyDeleteBTW, I think here is another typo:

whenever ab is a multiple of p then at least one of a and b must be a multiple of b.

Cheers.

Tomas

Right again, of course, and also fixed. Glad you enjoyed it.

ReplyDelete