Theorem 1 (Fermat’s Little Theorem).Let be a prime number and let be any integer. Then

In this post — *aimed at students taking an introductory course in number theory* — I want to explain the two proofs that I find to be the simplest. Neither is deep, but the second requires a tiny bit of group theory, whereas the first does not.

** 1. First proof: by binomial coefficients **

We shall take as given that the binomial coefficients , for are natural numbers. We also take as given the Binomial Theorem:

Fix a prime number .

Lemma 2.For , the binomial coefficient is divisible by .

*Proof:* Since is an integer, divides . Since is prime, and , it follows that and are coprime. By Euclid's Lemma, . Therefore is an integer.

Lemma 3.For integers we have .

*Proof:* Apply the Binomial Theorem and Lemma 1.

It’s now a simple matter to give the first proof of Fermat’s Little Theorem.

*Proof:* Let be a prime number, and let be the set of integers satisfying the conclusion of the theorem. If is odd, then . Otherwise, if then . Thus . Lemma 3 shows that is closed under addition. Therefore .

** 2. Second proof: by Lagrange’s Theorem **

The second strategy is reasonably simple. I think of it as the proof *‘by Lagrange’s Theorem’*, but that is perhaps unfair since its use of Bézout’s Lemma is just as key.

Let . Then there exist such that .

The proof goes by showing that the set

of -linear combinations of , is also the set of integer multiples of the greatest common divisor .

Let be a finite group, and let be a subgroup. Then .

The proof of Lagrange’s Theorem is a straightforward counting proof. One shows that each coset of has the same order, and that the cosets partition .

Okay! Using these two ingredients, we get the second proof of Fermat’s Little Theorem.

*Proof:* Let be a prime number. It is a consequence of Bézout’s Lemma that forms a group under multiplication modulo . Moreover it is a group of order . Thus, for each , we have .

Now let . If is divisible by , then certainly . Otherwise, there exists and such that . Then , as required.

To be perfectly honest, I find this second proof less satisfactory than the first. Relying on Bézout seems reasonable enough (we are interested in integer arithmetic after all), but relying on Lagrange is a bit silly. Lagrange’s Theorem is a statement about all finite groups, which is just far too strong for what we need.

However you get there, Fermat’s Little Theorem is an important result of elementary number theory.

]]>Ic eom mare þonne þes middangeard

læsse þonne hondwyrm leohtre þonne mona

swiftre þonne suna sæs me sind ealle

flodas on fæðmum on þes foldan bearm

grene wongas grundum ic hrine

helle underhnige heofonas oferstige

wuldres eðel wide ræce

ofer engla eard eorðan gefylle

ealne middangeard ond merestreamas

side mid me sylfum saga hwæt ic hatte

Find the prime factorization of .

**Solution.**

Write . As a first attempt we shall ask some straightforward questions:

- Is even? No.
- Is divisible by ? We use the “
*sum the digits*” trick, and find that — yes — is divisible by , but not divisible by . - Is divisible by ? No.

Etc. We can directly calculate for other small primes, and if we push on we will find a factor of .

Taking out these factors, namely and , we are still left with a large number:

How should we proceed? We seek a more systematic approach, or at least some sort of shortcut.

Fortunately, in this special case, such a short cut is available: we exploit the repetition of various strings of digits in the base- representation of . Let and let . Then , and . Rearranging, we get

It now remains to find the prime factorizations of , , and . The most attractive place to begin is with :

I don’t know of any particularly neat way of finding the factors and but it’s very easy by hand.

Next we consider :

The factor of is easy to find, but I don’t know a quick way to verify that is prime, other than trial division, by checking for prime factors up to .

Finally, we consider :

Again, I don’t know an easy way to find these factors, other than trial division, by checking for prime factors up to .

Putting all of this together, we have the prime factorization

As mentioned in the original post, perhaps the real challenge is to come up with a problem that retains the trickiness of this example, but avoids the cumbersome checking.

]]>Let be a graph, where we view as a set of -element subsets of . For , denote by the degree of . (Notation and terminology as in my graph theory lecture notes.)

Lemma 1..

It’s easy to let this little formula go by without much thought. It gets used in the proof of Mantel’s Theorem. But I want to bring it to the fore, as an example of a simple but perhaps opaque statement, which becomes clearer with a little abstraction.

Consider the following stronger statement.

Lemma 2.Let be any function. Then

*Proof:* Given , the term appears on the right hand side once for each edge of which is an endpoint.

Lemma 1 follows by taking .

Let’s see how Lemma 1 is used in a proof of Mantel’s Theorem.

Theorem 3 (Mantel’s Theorem).Suppose that is triangle-free, of order , and of size . Then

*Proof: * Since is triangle-free, for every we have . Therefore . Summing across edges and using Lemma 1, we have

Combining these things, we have

as required.

So there we have it. Lemma 1 has a straightforward application in this proof of Mantel’s Theorem.

Note also that the complete bipartite graph is triangle-free and has size .

Often bundled into the statement of Mantel’s Theorem is the additional claim:

Proposition 4.Let be triangle-free, of order , and of size . Then is isomorphic to .

*Proof:* If , then each inequality in the above proof of Mantel’s Theorem must be an equality. In particular, for each we have . Thus the neighbourhoods of and of form a partition of the vertex set of . It follows that is complete bipartite, and up to isomorphism the only such graph to have size is .

This short note is written for those taking my second-year analysis course.

Möbius transformations are brilliant. They are conformal (=angle preserving) maps from the *extended complex plane* to itself.

Definition 1 (The extended complex plane).Theextended complex planeis , i.e. together with an entirely new element, denoted .

For now, we won’t extend the domains of and to include ; so the extended complex plane has no more algebraic structure than the field of complex numbers.

Definition 2 (Möbius transformations).AMöbius transformationis a function such thatfor some such that .

The second and third cases in the definition above are important, but we specify a Möbius transformation simply by writing e.g.

and leave the other two cases implicit.

Nevertheless, such a Möbius transformation does not determine the quadruple , but does determine the tuple in projective space. For a trivial example: the identity map is a Möbius transformation, equal to .

Example 1.Consider the Möbius transformation defined by Note that really is a Möbius transformation since . A few example evaluations: , , and .

Proposition 3.The composition of two Möbius transformations is a Möbius transformation.

*Proof:* For , consider a Möbius transformation , where . The composition is which is a Möbius transformation since is not zero.

Proposition 4.Möbius transformations are bijections.

*Proof:* Let , with . We find the candidate for its inverse, and check it really is. Define

by

We compose:

the identity! The composition is the identity by the same calculation.

Definition 5 (Möbius group).The set of Möbius transformations, equipped with composition, forms theMöbius group, denoted .

We are able to represent Möbius transformations as matrices, in the following way.

Definition 6.Define the map

Naturally we wonder whether, under this representation of Möbius transformations as matrices, the composition of functions corresponds to the multiplication of matrices.

Proposition 7.is an isomorphism.

*Proof:* It is clear that is a bijection. For , consider defined by , where .

The composition is

and we have

which shows that is a homomorphism.

Proposition 8.Möbius transformations act -transitively on .

*Proof sketch:* To see this, let be three distinct elements of the extended complex plane. We wish to choose such that

satisfies

Equivalently, such that the three equations

hold. It suffices to satisfy which can indeed be accomplished.

]]>Along with colleagues all over the world, I will be spending much of the next few weeks and months at home. Working — teaching, research, etc. — but detached. How best should I make use of the time? I will try to post more puzzles for students, and little notes that may be of interest to students or fellow educators.

For now, here is a little one: how many configurations are there of a 2x2x2 Rubik’s cube?

For our purposes, a ‘configuration’ is simply any position into which a ‘solved’ cube may be transformed by any finite sequence of the allowed physical twists.

]]>Find the prime factorisation of 12318876808768112319.

Factorisation puzzles are familiar, and obviously the point is to minimise the use of computer assistance, by making ‘clever’ observations.

Once you see how to do it, you’ll notice that there is a still a non-trivial amount of checking to do. Thus, a meta-puzzle suggests itself: to find a natural number N, so that the problem of factorising N by hand contains lots of interesting and clever observations, and very little checking.

I don’t know how to solve this meta-puzzle!

]]>There are five films:

- Star Wars X,
- Avengers Rereunite,
- Frozen 3 – ice cold,
- Fish and chips – the inside story,
- Harry Potter and the Arithmancer of Andover.

There are three choices of snack:

- Basil popcorn,
- Chilli marshmallows,
- Vegan burger.There are four choices of drink:
- 8 down,
- Diet Cake,
- Popsi Minimum,
- Ginger coffee.

They don’t mind going to separate films, but Cally won’t miss Star Wars if either of the others sees it (because she’s worried about spoilers). Each of them wants a drink of their own, and a snack of their own, except that the bags of marshmallows must be shared between two of them.

How many configurations of cinema trip may our three friends choose?

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Question 1.Let be differentiable, with derivative . Suppose that . Show there exists such that .

Compare Question 1 with Rolle’s Theorem.

Question 2 (Darboux’s Theorem).Let be differentiable, with derivative . Show that has the intermediate value property.

Compare Question 2 with the Mean Value Theorem.

Question 3.Let be differentiable, with derivative . Show that there exists such that is continuous at .

This is very difficult! See this article. Nevertheless, I would like to expand on this argument in a subsequent post. Watch this space.

Question 4.Suppose that satisfies the intermediate value property. Need admit an anti-derivative?

Actually, this one is straightforward, so I’ll give the argument right away.

Proof.Certainly not. In fact we have just seen that every derivative (so, a function that admits an anti-derivative) is continuous somewhere. However, there are plenty of functions that satisfy the intermediate value property which are continuous nowhere. For example, there are functions whose graphs are dense in the plane.

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Remark.For an in-depth discussions of the continuity of derivatives, see this article.

We begin at the beginning, with a couple of definitions.

Definition.A function has theIVT Propertyif, for all real numbers with and there exists such that .

So, using this terminology, the Intermediate Value Theorem is the statement that a continuous function has the IVT Property.

Definition.Let and let . We say that is aweak limitof at if

for all there exists such that . Naturally, we say thathas a weak limit somewhereif there exists such that is a weak-limit of at .We say that is an

accumulation pointof a real sequence if for all there is an infinite set such that for all we have .

**The Questions**

Question 1.

Show the existence or non-existence of a function which has the IVT Property but which is nowhere continuous.

Question 2.

Prove or disprove the claim that every function has a weak limit somewhere.

Question 3.

Find a real sequence such that every real number is an accumulation point of .

Where should we start? Perhaps my favourite of the three is the first one: to show the existence or non-existence of a function with the IVT property but which is nowhere continuous. There are several approaches we might take, beginning with the following lemma.

Lemma 1.Suppose that a function has the property that, for all with , the restriction of to the interval is surjective onto . Then has the IVT property.

Hopefully the proof of this lemma is quite clear: the IVT property asks, of such a function restricted to such an interval, that it attain certain values. So if attains *all* real values, then the property is evidently satisfied.

Okay…. So what?

Lemma 2.Suppose that a function has the property that, for all with , the restriction of to the interval is surjective onto . Then is continuous nowhere.

Proof.Let be given, and consider, for example, . For any , by assumption restricted to the open interval is surjective onto . In particular, the image under of is not a subset of .

Ah-ha! Now it’s clear why we’re interested in these sorts of function. Such a `*locally surjective function*‘ would be a nowhere continuous function which nevertheless has the IVT property. The big question now is: does any locally surjective function exist?

Well, YES. Such functions do exist, and there are several ways to find them. First we give a slightly abstract construction, which might seem a little unsatisfactory.

Method 1.Recall that the cardinality of is , and the cardinality of is also . We may choose a bijection.

Also, let be the natural map sending each real number to its coset modulo , namely . Note that is surjective. Thus, the composition is also a surjection. Indeed, by the density of the rationals in the reals, each proper open interval contains a representative of each coset of in . That is, for each , there exists such that . A consequence of this is that the restriction to of is still surjective onto . Finally, the restriction of to is still surjective onto , which shows that is locally surjective, as required.

Wonderful, that was easy. But it wasn’t particularly constructive. Perhaps it felt a little unsatisfactory to rely on the cardinalities of these sets. Is there a better way?

That’s where I’ll leave it for the moment. To be continued!

In the meantime, try to construct a locally surjective function by using the decimal expansion of each real number.

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