# Balkan MO 2017 – Qs 1, 3 and 4

The UK is normally invited to participate as a guest team at the Balkan Mathematical Olympiad, an annual competition between eleven countries from South-Eastern Europe. I got to take part in Rhodes almost exactly ten years ago, and this year the competition was held in Ohrid, in Macedonia. There’s one paper, comprising four questions, normally one from each of the agreed olympiad topic areas, with 4.5 hours for students to address them. The contest was sat this morning, and I’m going to say quite a bit about the geometric Q2, and a little bit about Qs 1 and 3 also. In all cases, this discussion will include most of a solution, with some commentary, so don’t read these if you are planning to try the problems yourself.

I’m not saying anything about Q4, because I haven’t solved it. (Edit: I have solved it now, so will postpone Q2 until later today.)

Question One

Find all ordered pairs of positive integers (x,y) such that

$x^3+y^3=x^2+42xy+y^2.$

The first thought is that if either of x or y is ‘large’, then the LHS is bigger than the RHS, and so equality can’t hold. That is, there are only finitely many solutions. The smallest possible value of y is, naturally, 1, and substituting y=1 is convenient as then $y^2=y^3$, and it’s straightforward to derive $x=7$ as a solution.

Regarding the non-existence of large solutions, you can make this precise by factorising the LHS as

$(x+y)(x^2-xy+y^2) = x^2+42xy+y^2.$

There are 44 terms of degree two on the RHS, and one term of degree in the second bracket on the LHS. With a bit of AM-GM, you can see then that if $x+y>44$, you get a contradiction, as the LHS will be greater than the RHS. But that’s still a lot of possibilities to check.

It struck me that I could find ways to reduce the burden by reducing modulo various primes. 2, 3 and 7 all divide 42, and furthermore cubes are nice modulo 7 and squares are nice modulo 3, so maybe that would bring the number of possibilities down. But my instinct was that this wasn’t the right way to use the fact that we were solving over positive integers.

The second bracket in the factorisation looks enough like the RHS, that it’s worth exploring. If we move $x^2-xy+y^2$ from the right to the left, we get

$(x+y-1)(x^2-xy+y^2) = 43xy.$ (1.1)

Now it suddenly does look useful that we are solving over positive integers, because 43 is a prime, so has to appear as a factor somewhere on the LHS. But it’s generally quite restrictive that $x^2-xy+y^2 | 43xy$. This definitely looks like something that won’t hold often. If x and y are coprime, then certainly $x^2-xy+y^2$ and $y$ are coprime also. But actually if x and y have a non-trivial common factor d, we can divide both sides by $d^2$, and it still holds. Let’s write

$x=dm,\quad y=dn,\quad\text{where }d=\mathrm{gcd}(x,y).$

Then $m^2 -mn+n^2$ really does divide 43, since it is coprime to both m and n. This is now very restrictive indeed, since it requires that $m^2-mn+n^2$ be equal to 1 or 43. A square-sandwiching argument gives $m^2-mn+n^2=1$ iff $m=n=1$. 43 requires a little bit more work, with (at least as I did it) a few cases to check by hand, but again only has one solution, namely $m=7, n=1$ and vice versa.

We now need to add the common divisor d back into the mix. In the first case, (1.1) reduces to $(2d-1)=43$, which gives $(x,y)=(22,22)$. In the second case, after cancelling a couple of factors, (1.1) reduces to $(8d-1)=7$, from which $(x,y)=(7,1),(1,7)$ emerges, and these must be all the solutions.

The moral here seemed to be that divisibility was a stronger tool than case-reduction. But that was just this question. There are other examples where case-reduction is probably more useful than chasing divisibility.

Question Three

Find all functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that

$n+f(m) \,\big|\, f(n)+nf(m)$

for all $m,n\in\mathbb{N}$.

What would be useful here? There are two variables, and a function. It would be useful if we could reduce the number of variables, or the number of occurences of f. We can reduce the number of variables by taking m=n, to get

$n+f(n) \,\big|\, f(n) [1+n].$ (3.1)

From this, we might observe that $f(n)\equiv 1$ is a solution. Of course we could analyse this much more, but this doesn’t look like a 10/10 insight, so I tried other things first.

In general, the statement that $a\,|\,b$ also tells us that $a\,|\, b-ka$. That is, we can subtract arbitrary multiples of the divisor, and the result is still true. A recurring trope is that the original b is elegant, but an adjusted b-ka is useful. I don’t think we can do the latter, but by subtracting $n^2 +nf(m)$ from the problem statement, we get

$n+f(m) \,\big|\, n^2-f(n).$ (3.2)

There’s now no m on the RHS, but this relation has to hold for all m. One option is that $f(n)=n^2$ everywhere, then what we’ve deduced always holds since the RHS is zero. But if there’s a value of n for which $f(n)\ne n^2$, then (3.2) is a very useful statement. From now on, we assume this. Because then as we fix n and vary m, we need $n+f(m)$ to remain a divisor of the RHS, which is fixed, and so has finitely many divisors. So $f(m)$ takes only finitely many values, and in particular is bounded.

This ties to the observation that $f\equiv 1$ is a solution, which we made around (3.1), so let’s revisit that: (Note, there might be more elegant ways to finish from here, but this is what I did. Also note, n is no longer fixed as in previous paragraph.)

$n+f(n) \,\big|\, f(n) [1+n].$ (3.1)

Just to avoid confusion between the function itself, and one of the finite collection of values it might take, let’s say b is a value taken by f. So there are values of n for which

$n+b \,\big|\, b(1+n).$

By thinking about linear equations, you might be able to convince yourself that there are only finitely many solutions (in n) to this relation. There are certainly only finitely many solutions where LHS=RHS (well, at most one solution), and only finitely many where 2xLHS=RHS etc etc. But why do something complicated, when we can actually repeat the trick from the beginning, and subtract $b(n+b)$, to obtain

$n+b \,\big|\, b^2-b.$

For similar reasons to before, this is a great deduction, because it means if $b\ne 1$, then the RHS is positive, which means only finitely many n can satisfy this relation. Remember we’re trying to show that no n can satisfy this relation if $b\ne 1$, so this is definitely massive progress!

If any of what’s already happened looked like magic, I hope we can buy into the idea that subtracting multiples of the divisor from the RHS is the only tool we used, and that making the RHS fixed gives a lot of information about the LHS as the free variable varies. The final step is not magic either. We know that f is eventually 1. If you prefer “for large enough n, $f(n)=1$,” since all other values appear only finitely often. I could write this with quantifiers, but I don’t want to, because that makes it seem more complicated than it is. We genuinely don’t care when the last non-1 value appears.

Anyway, since we’ve deduced this, we absolutely have to substitute this into something we already have. Why not the original problem statement? Fix m, then for all large enough n

$n+f(m) \,\big|\, 1+nf(m).$ (3.3)

To emphasise, (3.3) has to hold for all large enough n. Is it possible that f(m)=2? Again, it’s easy to convince yourself not. But, yet again, why not use the approach we’ve used so profitably before to clear the RHS? In fact, we already did this, and called it (3.2), and we can make that work [3.4], but in this setting, because f(m) is fixed and we’re working with variable large n, it’s better to eliminate n, to get

$n+f(m)\,\big|\, f(m)^2-1,$

again for all large enough n. By the same size argument as before, this is totally impossible unless f(m)=1. Which means that in fact $f(m)=1$ for all m. Remember ages ago we assumed that f(n) was not $n^2$ everywhere, so this gives our two solutions: $f(n)=1,\, f(n)=n^2$.

Moral: choosing carefully which expression to work with can make life much more interesting later. Eliminating as many variables or difficult things from one side is a good choice. Playing with small values can help you understand the problem, but here you need to think about soft properties of the expression, in particular what happens when you take one variable large while holding another fixed.

[3.4] – if you do use the original approach, you get $n^2-1$ on the RHS. There’s then the temptation to kill the divisibility by taking n to be the integer in the middle of a large twin prime pair. Unfortunately, the existence of such an n is still just a conjecture

Question Four

(Statement copied from Art of Problem Solving. I’m unsure whether this is the exact wording given to the students in the contest.)

On a circular table sit n>2 students. First, each student has just one candy. At each step, each student chooses one of the following actions:

(A) Gives a candy to the student sitting on his left or to the student sitting on his right.

(B) Separates all its candies in two, possibly empty, sets and gives one set to the student sitting on his left and the other to the student sitting on his right.

At each step, students perform the actions they have chosen at the same time. A distribution of candy is called legitimate if it can occur after a finite number of steps.
Find the number of legitimate distributions.

My moral for this question is this: I’m glad I thought about this on the bus first. What I found hardest here was getting the right answer. My initial thoughts:

• Do I know how to calculate the total number of possibilities, irrespective of the algorithm? Fortunately yes I do. Marbles-in-urns = barriers between marbles on a line (maybe add one extra marble per urn first). [4.1]
• What happens if you just use technique a)? Well first you can get into trouble because what happens if you have zero sweets? But fine, let’s temporarily say you can have a negative number of sweets. If n is even, then there’s a clear parity situation developing, as if you colour the children red and blue alternately, at every stage you have n/2 sweets moving from red children to blue and vice versa, so actually the total number of sweets among the red children is constant through the process.
• What happens if you just use technique b)? This felt much more promising.
• Can you get all the sweets to one child? I considered looking at the child directly opposite (or almost-directly opposite) and ‘sweeping’ all the sweets away from them. It felt like this would work, except if for some parity reason we couldn’t prevent the final child having one (or more, but probably exactly one) sweets at the crucial moment when all the other sweets got passed to him.

Then I got home, and with some paper, I felt I could do all possibilities with n=5, and all but a few when n=6. My conjecture was that all are possible with n odd, and all are possible with n even, except those when none of the red kids or none of the kids get a sweet. I tried n=8, and there were a few more that I couldn’t construct, but this felt like my failure to be a computer rather than a big problem. Again there’s a trade-off between confirming your answer, and trying to prove it.

Claim: If n is even, you can’t achieve the configurations where either the red children or the blue children have no sweets.

Proof: Suppose you can. That means there’s a first time that all the sweets were on one colour. Call this time T. Without loss of generality, all the sweets are on red at T. Where could the sweets have been at time T-1? I claim they must all have been on blue, which contradicts minimality. Why? Because if at least one red child had at least one sweet, they must have passed at least one sweet to a blue neighbour.

Now it remains to give a construction for all other cases. In the end, my proof has two stages:

Step One: Given a configuration, in two steps, you can move a candy two places to the right, leaving everything else unchanged.

This is enough to settle the n odd case. For the even case, we need an extra step, which really corresponds to an initial phase of the construction.

Step Two: We can make some version of the ‘sweeping’ move precise, to end up in some configuration where the red number of children have any number of sweets except 0 or n.

Step one is not so hard. Realising that step one would be a useful tool to have was probably the one moment where I shifted from feeling like I hadn’t got into the problem to feeling that I’d mostly finished it. As ever in constructions, working out how to do a small local adjustment, which you plan to do lots of times to get a global effect, is great. (Think of how you solve a Rubik’s cube for example.)

Step two is notationally fiddly, and I would think very carefully before writing it up. In the end I didn’t use the sweeping move. Instead, with the observation that you can take an adjacent pair and continually swap their sweets it’s possible to set up an induction.

Actual morals: Observing the possibility to make a small change in a couple of moves (Step one above) was crucial. My original moral does still hold slightly. Writing lots of things down didn’t make life easier, and in the end the ideas on the bus were pretty much everything I needed.

[4.1] – one session to a group of 15 year olds is enough to teach you that the canon is always ‘marbles in urns’ never ‘balls’ nor ‘bags’, let alone both.

# The Top-to-Random Shuffle

This article is based on a talk I gave to the Maths Society at St Paul’s School on Monday. It may turn into a short series if I have time before I go to ALEA in Luminy near Marseille on Saturday.

My original plan had been to talk about riffle-shuffling, and some of the interesting mixing time themed results one can obtain. As a motivating example, I began by discussing the simpler top-to-random shuffle, and this proved sufficiently interesting to occupy the time I had been allowed (and mea culpa a bit more). It therefore seems worth writing a hopefully moderately accessible blog post on the subject. The aim of this post at least is to discuss the idea that repeatedly shuffling brings a pack of cards close to randomness. We have to settle on a definition of ‘close to randomness’, and find some ways to calculate this.

Suppose we are playing some bizarre card game where it is necessary that three cards labelled, uncontroversially, 1, 2 and 3 need to be placed in a random order. If we are organised, we can write down all the ways to do this in a list:

123, 132, 213, 231, 312, 321.

We want to select each of these with equal probability. We could for example use a dice. Most relevantly, even a computer as ancient as my laptop is very happy simulating a random choice from this set. (Now is not the time to talk about exactly how pseudo-random or otherwise this choice would be.)

Of course, when we play a sensible card game we have not three cards, but fifty-two. So the approach described above still works in theory, but no longer in practice, as the list of possible arrangements now has size 52!. Recall this is defined to be

$52!=1\times 2 \times\ldots \times 52.$

The reason we get this particular expression is that when we are choosing the first card, we have 52 possible choices. Then, regardless of what this first card actually is, there are precisely 51 cards left from which to choose the second card. So there are 52×51 ways to pick the first two cards in the arrangement, and so on, giving the answer. We can approximate how large 52! is by counting powers of ten rather crudely. It seems reasonable that it should be about $10^{65}$. Note that the number of atoms in the universe is *only* about $10^{80}$, so if we are going to write down this list, we better have very compact handwriting! But being serious, this number is way too large to realistically compute with, so we have to come up with some cleverer methods.

One way is to spread the cards out on a table then pick them up one at a time, ensuring at all times that the choice of card is uniform among those currently present, and not related to any of the past choices. This is relatively easy for a computer, but hard for a human, and certainly deeply tedious for anyone waiting to receive their hand!

So we seek a different approach, namely an algorithm for shuffling. Our aim is to introduce overall randomness by repeatedly applying some simple but random process. Note we have to be careful about our definition of ‘random’ here. The permutation 123456 is just as ‘random’ as the permutation 361524. That is, if they are fixed, then they are not random at all. Just because it is easier to decribe one of them verbally does not mean it is less random. For example, if I am trying to cheat at poker, then I might be able to if I knew the exact order of the cards in the pack before the dealer dealt. It wouldn’t matter what that order was. I would have to adjust my strategy based on the order, but it wouldn’t affect the fact that I had a massive advantage!

The shuffling algorithm to be discussed here is the top-to-random shuffle. Like all the best things in life, this does exactly what it says on the tin. At a given time, we remove the top card from the deck at present, and insert it at a randomly chosen point in the deck. This could be on the bottom, and it could also be back on the top. It feels like this possibility to remain constant can’t possibly help us, but later we will discuss why we need this.

In any case, it feels natural that if we keep applying this procedure, the arrangement of the deck should start to get more and more random, in the sense that knowing the original arrangement will tell us successively little about the current arrangement as time progresses. But we need to find a way to quantify this if we are to do any mathematics.

When we are talking about real numbers, it is fairly clear what it means if I say that the numbers 2, 1.1, 1.01, 1.001 and so on are getting closer and closer to 1. Indeed we can measure the distance along the number line between each term and 1, using the absolute difference. It is not so clear how to compute the distance between two probability distributions. Bearing in mind the fact that a distribution on the set of permutations of cards is defined to be a set of 52! probabilities that sum to 1, there will be a 52!-1 dimensional space (eg the plane is two-dimensional, the world is three-dimensional, *and so on* – whatever that means) where we have a nice distance formula already.

But this is not what we will choose to use. Rather we return to the cheating-at-poker analogy. Suppose I am playing some sort of game involving the pack of cards with my enemy. He or she thinks the deck is perfectly random, but I know the actual distribution. How big a profit can I make by exploiting this knowledge? This will be our measure of how far a distribution is from uniform. It turns out that this will coincide precisely with the formal definition of total variation distance, but that language belongs to a different level of rigour and is not relevant here.

What is relevant is an explanatory example. Suppose we start with the arrangement 12345678. We are now going to perform one iteration of the top-to-random shuffle. The outcome might, for example, be 23456178, if we insert the 1 between the 6 and the 7. Note there were 8 places for the card to go, so the probability of this particular outcome is 1/8. Now let’s see how I might use my knowledge of the distribution to my advantage. Suppose I suggest the bet that the bottom card is an 8. My enemy thinks the stack is uniformly randomly arranged, so the probability of this is 1/8. On the other hand, I know that the only way the 8 might disappear from the bottom is if I place the 1 under it, which happens with probability 1/8. So in fact, I know the probability of this event is 7/8, which gives me an advantage of 3/4. In fact, I could come up with bets that do even better than this, but they are less simple to describe verbally.

At what point do I lose this advantage? Well, we said that the probability that the 8 leaves the bottom of the stack is 1/8. And it will continue to be 1/8 on every turn where it is at the bottom. Recalling that the outcomes of successive shuffles are independent, note this is reminiscent of rolling a dice until a six comes up. The number of rolls required to get the six is an example of a geometric random variable. I don’t want to spoil S1 (or whichever module) by going into too much detail, but it turns out that if the probability of an event happening on a single go is p, then the average time we have to wait is 1/p. So 1/(1/8)=8 of course, and this is how long we typically have to wait before the bet I placed before becomes much less effective.

Now seems like a good time to stop talking about 8 cards and start talking about n cards. Obviously, in practice, we will want n to be 52. Anyway, by the same argument as before, it takes on average n iterations before the bottom card leaves the bottom. This is important, because after then, my bet that the bottom card is n is no longer so effective. However, I could equally place a bet that one of the bottom *two* cards is n.

So we consider how long it takes before n is no longer one of the bottom two cards. Well certainly we need to wait until it is no long *the* bottom card, which takes time n on average. Then, once it is second bottom, there is now a 2/n chance that we move the previously top card below it, so by the same argument as before, the time for this to happen is n/2 on average. If we want this effect to disappear, we have to wait until the original bottom card is in fact at the top of the pile for the first time, and by extending our previous argument, the average time for this is

$n+\frac{n}{2}+\frac{n}{3}+\ldots+\frac{n}{n-1}.$

Fortunately, we have tools for approximating this sort of sum, in particular integration, which is the practice of finding the area under certain curves. It turns out that the answer is roughly n log n. You can think of as log n a measure of the number of digits required to write out n. (This is not the exact definition but it will do for now. In any case, log n gets larger as n gets larger, but not very fast.) There’s a lot more about this in my previous post on the coupon collector problem, from a more technical point of view.

The next question will be to prove that it is actually quite well shuffled by this time, but that’s for another post. The other question to ask is whether this is satisfactory overall? For n=52, the number of operations we have to perform is about 230, which is fine for a computer, but deeply tedious for anyone sitting at a casino table waiting for the next hand. So next time we’ll talk about the riffle shuffle, which seems to introduce a lot of randomness in each go, but we’ll also see that we have to be careful, because the randomness may not be as great as our intuition suggests.