Lecture 11
In general, if we have positive integers and , with , we can start defining a sequence as follows:
,
,
is the remainder upon dividing by :
This is a decreasing sequence, and eventually we will get for some ; we can’t divide by zero, so we end the sequence there.
We then have
Let’s write for this.
Now, we have , so , so we can write as a linear combination of and .
We have , so , so substituting in we can write as a linear combination of and .
Proceeding in this way, we end up with as a linear combination of and : in other words, of and .
We’ve proved the following:
Proposition
[Bezout’s Lemma] Let and be two integers with . Then there are integers and such that .
In fact, slightly more is true:
Proposition
Let and be two integers with . Then, for an integer , we can write in the form if and only if .
Proof
The “if” part: We must prove that, if , then we can write as a linear combination of and .
However, since , we can write for some . Also, by the above Proposition we can write for some and . But then as required.
The “only if” part: We must prove that if , then . But, since we have and , and hence also and , and therefore as required.
The fundamental theorem of arithmetic
We’ll go on now and describe three uses of this result. Firstly, we return to the question of unique factorisation into primes. Of course we’ve proved that every positive integer can be written as a product of primes. The question is, can every positive integer be written as a product of primes in only one way?
Of course, we should be careful to say what we mean by “only one way”. We certainly do have: Clearly, what we mean is that every positive integer can be written as a product of primes in only one way, where reordering doesn’t count as different. Or, more precisely, that any two ways of writing a positive integer as a product of primes differ only by reordering. Mathematicians say, “in only one way, up to reordering”.
So the question we ask ourselves is (for example) why we can’t have (I promise you that all six of those numbers are prime).
One wants to say something like “as the right-hand side is clearly divisible by , the left-hand side must be divisible by too, but there isn’t a listed among the primes on the left”.
But if we have , why must we have either or ? It wouldn’t be true if weren’t a prime. But this is true for primes!
Proposition
Let be a prime, and and be integers. Then, if , then or .
This result is not only not obvious, we should expect it to be difficult. The definition of “ being prime” talks about what things divide . But this result says something about what things divides, which is completely unrelated.
Proof
Suppose that , and consider . Since , we either have or .
If , then as , we have .
If , however, then by Bezout’s Lemma, we know that there are integers and such that . Now suppose we multiply both sides by ; we get .
Clearly , and also we have since we are supposing that . Hence , so , as needed.
The second part of this proof can in fact be used to show that, for any integers , and , that if and , then .
We can also boost it to a result about a product of lots of terms:
Proposition
Let be a prime and let be integers. Then if , then for some .
This is an easy induction argument using above.
Now, equipped with that tricky result, we’re ready to prove the main result of this section:
Theorem
[Fundamental Theorem of Arithmetic] Any positive integer can be written as a product of primes in exactly one way, up to reordering.
Proof
We have shown that any positive integer can be written as a product of primes. We need to show that this expression is unique. We’ll prove it by contradiction.
Suppose not: there is a number with two genuinely different prime factorisations and . We can suppose that the ’s and the ’s have nothing in common (if , then we can cancel them out and use , which is a smaller example).
Now, that means that is different to all of .
We have , since . But then we also have . But by our previous result, this means that for some . But, by the definition of being a prime number, that means that , which we said didn’t happen: that gives us our contradiction.
Linear diophantine equations
A diophantine equation is an equation where we’re interested in solutions with the variables lying in or . They’re named after the ancient Greek mathematician Diophantus of Alexandria.
An example of a diophantine equation is the Fermat equation for exponent 7:
If we were interested in solutions to this equation over , the story would be really, really simple: we could take any and any we wanted and then just take The Fermat equation becomes more interesting because of our inability to reliably take th roots in or : which and can we take for which this recipe works?
While they’re much easier, a similar thing is true of linear diophantine equations: equations of the form where , and are integer constants.
Consider, for example, the equation .
This equation would be simple if we cared about real solutions: we could take any we like and then just take . However, because we can’t do division reliably in , this recipe is not very helpful: how do we know which will give us an integer ?