It’s illustrative to move from the graphical definition of a Latin square to something more symbolic. A Latin square on an index set *I* is a function *f* assigning each element of an element such that for each fixed , and realize each value in exactly once. Blurring the distinction between and , we can recast the above definition in the following way.

*Definition.* A *Latin square* on a (finite) index set *I* is a subset *S* of such that, for any fixed , each of the incomplete triples can be completed to an element of in exactly one way. (That is, a subset of with size whose projection onto any pair of coordinates is exactly .)

Notice that “function-ness” and the conditions on the rows and columns are absorbed into one pleasantly symmetric condition. This explains the mysterious fourth paragraph of this Not About Apples post. (This definition also generalizes gracefully to orthogonal Latin squares; a list of *k* orthogonal squares corresponds naturally to a subset of with size whose projection onto any pair of coordinates is exactly .) This symmetry also means that acts naturally on the set of Latin squares on a particular index set by permuting the coordinates. This led me to think about the following definition.

*Definition*. A *supersymmetric Latin square* (*ssLs*) is a Latin square which (viewed a subset of ) is closed under permutation of coordinates.

That is, a supersymmetric Latin square is in an orbit by itself under the -action.

*Example.*

There are three supersymmetric Latin squares on {0,1,2}.

021 102 210 210 021 102 102 210 021

For ordinary Latin squares, there is another kind of symmetry. Rows can be permuted, columns can be permuted, and the symbols themselves can be permuted. That is, the symmetric group on the index set acts naturally on the set of Latin squares in three ways, by acting on the first, second, or third coordinate of the triples. This justifies the common convention of assuming without loss of generality that a Latin square has the symbols listed in order across the first row and down the first column; since each of these actions is free (though the composite action of is not), we have an instant proof that divides the number of Latin squares on a set of *n* (and if you are a little careful, you can improve this to for almost no extra work).

This doesn’t work for supersymmetric Latin squares, though; these actions of are not compatible with the action of in the definition of supersymmetry. There does, however, exist a natural action of on the set of supersymmetric Latin squares on , where a permutation of acts on all permutation . In further contrast with the ordinary case, this action is not free; it’s not even faithful.

**Example.**

The first ssLs listed above is stabilized by all of . The latter two are stabilized by the even permutations; odd permutations interchange them.

I call two supersymmetric Latin squares on *I equivalent* if they are in the same orbit of this -action. Likewise, ssls on *I* is equivalent to an ssls on *J* if the latter is the image of the former under some bijection acting on every component of every triple. I’m interested in classifying supersymmetric Latin squares of size *n*, up to equivalence.

~~For , it’s not hard to check that there are no examples. ~~ *(Edit: as noted in the comments, that’s just plain not true.)* For there are the two already given. Call them in the order they are listed in the first example.

There are infinitely many supersymmetric Latin squares. Let me briefly explain the construction of all the ones I know.

There is a natural product on Latin squares. If *S* (resp. *S’*) is a Latin square on *I* (resp. *I’*), then we define , a Latin square on , by . I leave to the reader the verification that this is fact a Latin square, and that it is supersymmetric if *S* and *S’* are.

Since we have two ssLs’s of size 3, we can construct ssLs’s of size 9. There are four possibilities to consider: . All will be supersymmetric, of course, but they cannot all be inequivalent. The middle two are certainly equivalent, since the product operation is commutative up to equivalence. Now we can just write out . (In order to write the squares in my preferred way, I have to choose a bijection between and ; here I use the one provided by base 3 notation.)

021687354 102768435 435102768 210876543 021687354 354021687 102768435 210876543 543210876 687354021 768435102 102768435 876543210 687354021 021687354 768435102 876543210 210876543 354021687 435102768 768435102 543210876 354021687 687354021 435102768 543210876 876543210

It’s not obvious, but the second and third squares are actually equivalent under e.g. the permutation .

The same approach enables us to construct two supersymmetric Latin squares and of size for . Any mixed product of ‘s and ‘s is equivalent to , so we have found all the ssLs’s which arise as products of the ones we know. These two are inequivalent, because contains all triples of the form and contains no such triples.

**Questions.**

- Do supersymmetric Latin squares exist of any size which is not a power of 3?
- For size any positive power of 3, we have found two inequivalent ssLs’s. Is this a complete list?
- We could also consider less strict symmetry conditions, requiring only that the Latin square be invariant under some subgroup of . If the subgroup is generated by a transposition, then this boils down to considering Latin squares which are symmetric (in the sense of symmetric matrices). But what about requiring only -symmetry, that the square be invariant under cyclic shifts. (I haven’t really thought about this yet, but it seems potentially interesting.)

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Speaking of the gamma function, I was teaching a probability class, and it so happened that integrals of the form came up frequently in the examples and homework. I overheard two of my students remarking on this phenomenon before class.

“I’ve noticed that integrals like that always come out to factorials,” said the first guy, “1, 2, 6, and so on.”

So now he has my undivided attention. This kid wasn’t even a math major, and he picked up on the pattern. He’s clearly not done talking, and I want to know what his next piece of insight is going to be.

“It’s kinda useful,” he went on, “because when I need to figure out a big factorial I don’t know, like or something, I can just write down the integral and do it on my calculator.”

To this day, I consider it one of my greatest accomplishments as an educator that I did not laugh nor spray the coffee I was drinking out my nose.

“Why don’t you just use the factorial button?” asked his friend.

“There’s a factorial button?”

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Suppose a set of positive integers contains the number 4 and is closed under the maps and . Must the set be all of $\mathbb{Z}^+$?

The motivation comes from that game that almost everyone who likes to play with numbers has played, the four fours game, where you try to express various numbers in terms of four 4s and whatever operations you may wish.

This is the corresponding game with only one 4; accordingly, we only use unary operations, and only those which return integers. What numbers can be expressed by starting with a 4, using factorials to go up, floored square roots to come down, going back and forth between these as desired. Mel Hochster has verified by computer that all positive integers up to 1000 are so expressible and conjectures that *all* positive integers are.

But how to prove it. He says that he asked Conway, who said something like “It’s hopeless.” Not encouraging, sure, but I’m a person who spends time working on the Collatz problem, about which Erdős famously said “Mathematics is not ready for such problems.”

Apparently I have no idea when I’m out of my depth. But it seems an interesting puzzle. A good place to start seems to be a rather easier question.

Let *n* be a positive integer. Let be the minimal *m* such that for some integer . Can we get any nontrivial estimates on how grows?

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So last weekend at MathFest, I was pleased to learn of the following inequality involving binomial coefficients.

Let be positive integers. .

Apparently this inequality has been proven, but only by a lengthy and uninspiring computation involving estimating gamma functions. There is not yet a combinatorial proof. What do I mean by a combinatorial proof of an inequality? I mean an interpretation of each of the nested binomial coefficients as the number of objects in some set, and then an explicit bijection of one set with a proper subset of the other.

In this case, I am thinking of as counting the number of ways to choose *a* different (not necessary disjoint) *b*-element subsets of a set of *c* elements. For definiteness I can take the set of *c* elements to be , list the elements of a subset in increasing order, and list the subsets in lexicographical order. This suggests the following combinatorial formulation of the inequality.

Let be positive integers. Let *S* be the set of matrices with entries in . Let consist of those matrices in *S* for which the elements in each row are strictly increasing, and the rows are lexicographically strictly increasing. Let consist of those matrices in *S* for which the elements in each columns are strictly increasing, and the columns are lexicographically strictly increasing. Find an explicit bijection of of some proper subset of .

Haven’t cracked it yet, but I’m enjoying the attempt.

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**Theorem: **If are nonzero integers, at most one of which is negative, then the number of inequivalent occurrences of three mutually tangent spheres of curvature is the same as the number of algebraic integers in with norm , up to multiplication by a unit. The bijection between algebraic integers with that norm and triples of spheres is natural and explicit.

There is a nice way to compute this number. Define an arithmetic function on the positive integers as follows. If or , then if is even and if is odd; if , then ; for all . Extend to all positive integers by multiplicativity. Then .

**Theorem: **If are nonzero integers, at most one of which is negative, then the number of inequivalent occurrences of tangent spheres of curvature is given by the formula . Here is the number of solutions to the congruence , is the number of solutions to the congruence , and is 1 if *and* , otherwise 0.

(This strange formula comes from an application of Burnside’s Counting Lemma; as always in these problems, the trickiest part is keeping track of which solutions correspond to the same packing.)

The next natural problem in the progression, counting the total number of inequivalent occurrences of a given curvature in integer sphere packings, remains resistant to my current approach.

In the next week, I should have the corresponding result for counting occurrences of a given -tuple of -dimensional hyperspheres. If I’m lucky, my 2- and 3-dimensional techniques will generalize to let me count occurrences of -tuples in dimensions, but I can’t be sure of that part yet. Beyond that, I currently have no idea how to proceed.

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You can get the talk slides here. (Just right-click and select “Save”.)

If you want more in-depth information on Apollonian Circle Packings, the best place to start is probably a sequence of five papers. (If you want to really know everything, I recommend reading them in the listed order; if your interests are more strictly number-theoretic, then perhaps start with the fourth and jump back to the earlier papers on an as-needed-basis.) The first four are by Graham, Lagarias, Mallows, Wilks, and Yan. The fifth is by Erikkson and Lagarias.

- Geometric Group Theory I
- Geometric Group Theory II
- Geometric Group Theory III
- Number Theory I
- Number Theory II

A reference for Elena Fuchs’ result is here.

Sarnak’s letter to Lagarias (in which is proved the “twin prime conjecture”) is here.

As Morpheus says, time is always against us. These slides were written for a 15-minute talk, and I could easily have given two 60-minute talks on this topic, and a third on the generalizations I’ve been playing with most recently.

This talk was part of a special session on open and accessible problems in number theory and algebra, and I tailored it accordingly. You’ll notice that I’ve written a lot more about questions I *haven’t* answered than questions I *have*. My own discoveries were present only “obliquely”. Upon my return to Michigan, I hope to complement these slides with some more posts including material from my papers-in-progress and my freshest thoughts on this subject. So there’s more on the way.

Most importantly (and those of you who saw my talk will know this already), if any part of this interests or intrigues you, **contact me**. Use comments here, email me, find your way to the Cafe Aroma in Fenton, whatever. Graduates and undergraduates, I’m talking to you. There is a lot of accessible stuff here at a lot of levels and with a lot of flavors. Want to collaborate? I am friendly, and I will always work with students. Just want some more information? Please ask; if I don’t know the answer, I will find someone who does.

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**Theorem.**

Call a rectangle *semi-integral* if it has at least one side of integer length. Suppose that a rectangle is decomposed into a finite collection of smaller rectangles, each of which is semi-integral. Then the original rectangle is semi-integral.

Of course this would be trivial if we were talking about “fully integral” rectangles with both sides of integer length, but as it stands it is far from obvious. It is certainly possible to decompose a rectangle into a collection of semi-integral rectangles in such a way that some rectangles have non-integral length and others have non-integral width, and based purely on geometric concerns, it seems plausible that a counterexample might exist. Amazingly (to me), the simplest proof I know is this theorem is through integral calculus (and it is very simple indeed)!

**Proof of Theorem.**

Orient the rectangles so that the sides are parallel to the coordinate axes, so that every rectangle involved has the form . For any real numbers , consider the function . The integral of this function over a rectangle, is relatively easy to compute: By the periodicity of the sine, this integral will vanish for every $\alpha,\beta$ if and only if the rectangle is semi-integral. If integrates to 0 on every subrectangle in the decomposition, then integrates to 0 on the large rectangle. .

Note that the proof immediately generalizes to higher dimensions! If an *n*-dimensional box is decomposed into a finite union of sub-boxes, each of which has at least one integral side length, then the original box has at least one integral side length.

The mathematician Alain Connes has publicly said that one cannot truly understand the integers if one does not understand this problem.

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Make $\mathbb{Z}$ into a topological space by taking as a neighborhood basis the set of arithmetic progressions $A(a,d):=\{a+nd:n\in\mathbb{Z}\}$ (where , of course). It’s not hard to check that this really is a topology, and that it has the following interesting properties.

- The union of finitely many arithmetic progressions is both closed and open. (Look modulo the gcd of the various differences in the progressions; this set and its complement are each unions of residue classes.)
- Any open set which is not empty is infinite. (Obvious, since the neighborhoods are all infinite.)
- The set is open; by 1, it is also closed if there are only finitely many primes.
- But is finite, so is not open, and is not closed.

That’s the whole proof. Once you introduce the topology, the theorem practically proves itself!

(If someone has written about this topology and given it an official name, then I don’t know about it; “Furstenburg topology” seems as good a name as any for now.)

It is surprising to me that we don’t seen this topology (or its natural generalization to general number rings) mentioned more often. Sure, Furstenburg’s proof isn’t so much a new idea about primes as a convenient distillation of some key bits of elementary number theory, but there are lots of places in number theory where the discussion can be phrased in terms of this topology. Perhaps there is something to gain from this.

For example, if is a set of integers, then the Furstenburg closure of is the set of all integers such that, for every modulus , there is some such that .

**Exponential Diophantine equations.** For example, in recent threads on MathOverflow (beginning here, then moving here), Kevin Buzzard asks whether, if you want to know whether has solutions in natural numbers , it is always possible to either give a solution or give some modulus where the equation is “obviously” impossible. This is a good question, because it is a routine thing to enumerate the possible residues of , or any similar exponential Diophantine expression in any given modulus; it would be good to know whether the process of successively trying moduli and see what’s impossible would eventually rule out any given for which there is no solution. If the answer to his question is no, then one wonders what kind of proof methods would be required to prove that has no solutions in a case where it has solutions modulo for every . In other words, is .

(As noted in one of the comments on the second linked-to MathOverflow threads, it is a nice theorem that for any integer , the set is Furstenburg closed.)

In some very fluffy sense, a basic form of the Hasse principle is the assertion that the sets of numbers expressible by certain types of Diophantine expressions are Furstenburg closed**.
**

**Apollonian circle packings.** A primary open problem in integer Apollonian circle packings is about the set of curvatures that occur in a given packing. Fix a primitive packing *P*, and let *S* be the set of positive curvatures appearing in *P*. Then it’s well-known that the elements of *S* miss many residue classes modulo for some depending on *P*. (From a modular arithmetic point of view, this is in some sense the only obstruction, since it is also known that *S* intersects every residue class modulo *M* if *M* is coprime to 6.) The big question is, *does S contain all but finitely many of the numbers not ruled out by this modular arithmetic obstruction?*

In the context of the Furstenburg topology, the question has a nice summary. *Is S Furstenburg-relatively-closed in some set ?*

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Pictures such as these are constructed by inscribing a triple of three circles inside a larger circle, inscribing a circle in each lune created, and iterating the process. The numbers labelling the circles are the curvatures (1/radius). (The exterior circle has radius 1, but because it is exterior we say that it’s curvature is -1 by convention.) To a number theorist, the amazing fact is that if the four circles we begin with have integral curvature (in the figure, the starting circles have curvature -1, 2, 2, 3), or indeed if any four mutually tangent circles in such a packing have integral curvature, **all** the rest of the circles automatically have integral curvature. Such pictures are called integer Apollonian circle packings (IACPs).

One question I like to think about, the one that got me into the subject, is to think about which numbers appear in IACPs. That is, think about all PACPs at once. How many times does a given curvature appear? What pairs of numbers can appear? How often? All of my counting is up to symmetry; the picture above, for example, accounts for only one occurrence of 6, not four, and only one occurrence of the pair (2,3).

It turns out that answers to these counting problems can be unexpectedly elegant.

(Note: as the “Out[56]” which I clumsily left in the picture suggests, I generated this diagram and all my other Apollonian pictures using Mathematica 7; please contact me if you’re interested in methods for constructing pictures of this sort.)

Four numbers are the curvatures of four mutually tangent circles iff they satisfy the Descartes equation .

Given a triple of mutually tangent circles with curvatures , the possible curvatures of a fourth circle are the roots of the above equation, viewed as a quadratic in ; in general there are two possibilities, . In particular, their sum is . If we know one curvature , then the “other” possibility is .

A nice upshot of this is that, if we know the curvatures of four mutually tangent circles in a packing, we can compute all the rest using only addition and subtraction!

Let be fixed integers, at least one positive, and let us count the occurrences of adjacent circles of curvatures .

The trick is to identify any occurrence of the pair of with the doubly-infinite chain of circles/numbers tangent to both. For the which appears in the above picture, the corresponding chain is . Can we characterize the chains attached to a pair ?

It’s easy to prove, using the equations in section 0, that if is in such a chain, then all the curvatures in the chain have the form for . The key is to realize that, for fixed and varying , all of these quadratic functions have the same leading coefficient and the same minimum .

Then what we are really counting is integer sequences of the form , where , where without loss of generality we can take (choices of which differ only by sign lead to the same sequence in reverse order). Such must have the form , where , and .

In other words, there is a natural bijection between occurrences of in IACPs and square roots of modulo .

What we’ve said so far applies for any pair of integers . In the special case where are relatively prime, we can actually say a bit more, or at least a more aesthetically pleasing version of what we already said. Since, modulo , we have . Since is invertible modulo , the square roots of correspond to the square roots of . That is, if , then the number of occurrences of the pair up to symmetry is the number of square roots of modulo .

Let’s look harder at the specific packing pictured above. Since we only want to consider things up to symmetry, let’s just consider one quarter of the circle, from one of the 2’s to one of the 3’s. What numbers appear? In between 2 and 3 is 6. In between 2 and 6 is 11; in between 3 and 6 is 14. Etc. Perhaps motivated by the result of the previous section, note that each of these numbers is 1 more than the sum of squares.

If you have seen Farey fractions or the Stern-Brocot tree, you probably have a guess now as to what’s going on. The circle 2 corresponds to the fraction (or, if you prefer, the pair ), the circle 3 corresponds to , and between circles and we inscribe a circle corresponding to . (A circle corresponding to has curvature .)

Once guessed, this is easy to prove by induction. If appear as a mutually tangent quadruple of curvatures with “between” the other circles of positive curvature, then the circle in between and has curvature , i.e. which is just what we hoped. Likewise the circle between and has curvature . This completes the induction.

Notice that section 2 does not depend logically on section 1. Also, the diagram shown is the only integral packing in which the external circle has radius 1, so the only occurrences of pairs with appear in this diagram. Combining all this, we have two independent descriptions of which numbers appear on the circumference of the packing shown and how often (up to symmetry).

- Section 1 tells us that there is one occurrence of for every pair of square roots of modulo .
- Section 2 tells us that there is one occurrence of for every expression where is a fraction in lowest terms in the interval .

Rather surprisingly, we have obtained an alternate proof of the classical fact that the number of representations of an integer in the form with and is equal to the number of (pairs of ) square roots of modulo !

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That’s the idea behind Cap’s Whiteboard. This blog is offered as your way to drop in on me and hear about the mathematics that I’m learning, that I’m thinking about, that I’m doing. Think of these posts as what might happen if you asked me about my office blackboard, or one of my home whiteboards. Expect some combination of my latest research, problems I’m chewing on, particularly interesting articles I’ve read, and my idiosyncratic perspectives on “familiar” mathematics.

This is intended as blog written by a mathematician for other mathematicians, and I’m not making any promises that individual posts will be self-contained. Some stuff will be suitable for strong math majors, some for graduates, and likely some will only be for number theorists. Please feel free and encouraged, though, to ask me (by email or, preferably, in comments) for references if you want some background on anything you see here.

Posts are on no set schedule, and the amount of time between posts will probably have a high variance, depending on what else is going on for me mathematically. I expect to average about a significant post a week, not counting links and the like.

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