BMO2 2018

The second round of the British Mathematical Olympiad was taken yesterday by the 100 or so top scoring eligible participants from the first round, as well as some open entries. Qualifying for BMO2 is worth celebrating in its own right. The goal of the setters is to find the sweet spot of difficult but stimulating for the eligible participants, which ultimately means it’s likely to be the most challenging exam many of the candidates sit while in high school, at least in mathematics.

I know that lots of students view BMO2 as something actively worth preparing for. As with everything, this is a good attitude in moderation. Part of the goal in writing about the questions at such length (and in particular not just presenting direct solutions) is because I think at this level it’s particularly easy to devote more time than needed to preparation, and use it poorly.

All these questions could be solved by able children. In fact, each could be solved by able children in less than an hour. You definitely count as an able child if you qualified or if your teacher allowed you to make an open entry! Others count too naturally. But most candidates won’t in fact solve all the questions, and many won’t solve any. And I think candidates often come up with the wrong reasons why they didn’t solve problems. “I didn’t know the right theorems” is very very rarely the reason. Olympiad problems have standard themes and recurring tropes, but the task is not to look at the problem and decide that it is an example of Olympiad technique #371. The task is actually to have as many ideas as possible, and eliminate the ones that don’t work as quickly as possible.

The best way to realise that an idea works is to solve the problem immediately. For the majority of occasions when we’re not lucky enough for that to happen, the second-best way to realise that an idea works is to see that it makes the problem look a bit more like something familiar. Conversely, the best way to realise that an idea doesn’t work is to observe that if it worked it would solve a stronger but false problem too. (Eg Fermat’s Last Theorem *does* have solutions over the reals…) The second-best way to realise that an idea doesn’t work is to have the confidence that you’ve tried it enough and you’ve only made the problem harder, or less familiar.

Both of these second-best ideas do require a bit of experience, but I will try to explain why none of the ideas I needed for various solutions this year required any knowledge beyond the school syllabus, some similarities to recent BMOs, and a small bit of creativity.

As usual, the caveat that these are not really solutions, and certainly not official solutions, but they are close enough to spoil the problems for anyone who hasn’t tried them by themselves already. Of course, the copyright for the problems is held by BMOS, and reproduced here with permission.

Question One

I wrote this question. Perhaps as a focal point of the renaissance of my interest in geometry, or at least my interest in teaching geometry, I have quite a lot to say about the problem, its solutions, its origin story, the use of directed angles, the non-use of coordinate methods and so on. In an ideal world I would write a book about this sort of thing, but for now, a long and separate post is the answer.

This will be available once I’ve successfully de-flooded my apartment.

Question Two

I also wrote this problem, though I feel it’s only fair to show the version I submitted to the BMO committee. All the credit for the magical statement that appears above lies with them. There is a less magical origin story as well, but hopefully with some interesting combinatorial probability, which is postponed until the end of this post.One quick observation is that in my version Joe / Hatter gets to keep going forever. As we shall see, all the business happens in the first N steps, but a priori one doesn’t know that, and in my version it forces you to strategise slightly differently for Neel / Alice. In the competition version, we know Alice is done as soon as she visits a place for a second time, but not in the original. So in the original we only have to consider ‘avoid one place’ rather than the multiple possibilities now of ‘avoid one place’ or ‘visit a place again’.

But I think the best idea is to get Alice to avoid one particular place c\not\equiv 0 whenever possible. At all times she has two possible options for where to go next, lets say b_k+a_k, b_k-a_k in the language of the original statement. We lose nothing by assuming -N/2 < a_k\le N/2, and certainly it would be ridiculous for Joe / Hatter ever to choose a_k=0. The only time Alice’s strategy doesn’t work is when both of these are congruent to c, which implies N\,|\, 2a_k, and thus we must have N= 2a_k. In other words, Alice’s strategy will always work if N is odd.

I think it’s really worth noticing that the previous argument is weak. We certainly did not show that N must be odd for Alice to win. We showed that Alice can avoid a congruence class modulo an odd integer. We didn’t really need that odd integer to be N for this to work. In particular, if N has an odd factor p (say a prime), then the same argument works to show that we can avoid visiting any site with label congruent to 1 modulo p.

It’s actually very slightly more complicated. In the original argument, we didn’t need to use any property of b_k. But obviously here, if b_k\equiv 1 modulo p and p\,|\,a_k, then certainly b_{k+1}\equiv 1 modulo p. So we have to prove instead that Alice can ensure she never ‘visits 1 modulo p for the first time’. Which is fine, by the same argument.

So, we’ve shown that Neel / Alice wins if N is odd, or has an odd factor. The only values that remain are powers of 2. I should confess that I was genuinely a little surprised that Joe / Hatter wins in the power of 2 case. You can find a construction fairly easily for N=2 and N=4, but I suspected that might be a facet of small numbers. Why? Because it still felt we could avoid a particular site. In order for Alice’s strategy to fail, we have to end up exactly opposite the particular site at exactly the time when the next a_k=N/2, and so maybe we could try to avoid that second site as well, and so on backwards?

But that turned out to be a good example of something that got very complicated quite quickly with little insight. And, as discussed at the beginning, that’s often a sign in a competition problem that your idea isn’t so good. (Obviously, when composing a problem, that’s no guarantee at all. Sometimes things are true but no good ideas work.) So we want other ideas. Note that for N=4, the sequence (2,1,2) works for Joe / Hatter, because that forces Alice / Neel to visit either (0,2,1,3) or (0,2,3,1). In particular, this strategy gave Alice no control on the first step nor the last step, and the consequence is that we force her to visit the evens first, then transfer to an odd, and then force her to visit the other odd.

We might play around with N=8, or we might proceed directly to a general extension. If we have a Joe / Hatter strategy for N, then by doubling all the a_ks, we have a strategy for 2N which visits all the even sites in the first N steps. But then we can move to an odd site eg by taking a_N=1. Just as in the N=4 case, it doesn’t matter which odd site we start from, since if we again double all the a_ks, we will visit all the other odd sites. This gives us an inductive construction of a strategy for powers of two. To check it’s understood, the sequence for N=8 is (4,2,4,1,4,2,4).

Although we don’t use it, note that this strategy takes Alice on a tour of sites described by decreasing order of largest power of two dividing the label of the site.

Question Three

I have a theory that the average marks on Q1, Q2 and Q3 on this year’s paper will be in ascending order rather than, as one might expect, descending order. I think my theory will fail because it’s an unavoidable fact of life that in any exam, candidates normally start at the beginning, and don’t move to the middle until making earlier progress. But I think that’s the only reason my theory will fail.

Like kitchen cleanliness or children’s character flaws, it’s hard to compare one’s own problem proposals with others’ rationally. But I felt that, allowing for general levels of geometry non-preference, Q3 was more approachable than Q2, especially to any candidate who’d prepared by looking at some past papers.

I’m in no way a number theorist, but I know three or four common themes when one is asked to prove that a certain sequence contains no squares, or almost no squares. [3a]

  • Number theoretic properties of the sequence of squares. Squares cannot be 3 modulo 4 for example. They also cannot be 2 modulo 4, and thus they also cannot be 2^{k-1} modulo 2^k for any even k. This first observation was essentially the body of most solutions to Q4 of BMO1 2016, among many others.
  • Soft properties of the sequence of squares. The sequence of squares grows quadratically. Sometimes we can show a quadratic sequence will have no overlap with some other sequence for basic reasons. This is especially common if the second sequence is also quadratic or similar. For example, the expression n^2+3n-4 is typically not a square because

(n+1)^2 = n^2+2n+1 < n^2 + 3n - 4 < n^2+4n+4 = (n+2)^2,

  • when n is large. In fact the right hand inequality is always true, and the left hand inequality is true for n\ge 6, which doesn’t leave too many cases to check (and n=5 does actually give a square). This type of argument has been quite common on BMO recently, directly on Q1 of BMO1 2011 and also Q3 of BMO1 2016. An example in a more abstract setting is Q3 of Balkan MO 2007, which I greatly enjoyed at the time…
  • Number theoretic properties of the definition of a square. A square is the product of an integer with itself, and so if we want the product of two or more integers to be a square, then this imposes conditions on the shared factors of the two integers. I’ll cite some examples shortly.
  • Huge theorems. Some old paper which I encountered as a child asked us to find all solutions to x^2-1=2^y. Or similar – I can’t find it now – but Q2 of BMO2 2006 is close enough to the sensible approach to the problem. I think it’s more helpful to think about this as proving that a particular sequence rarely includes powers of two than that a particular sequence rarely includes squares. But either way, one could in principle use the Catalan conjecture, which controls all non-trivial solutions to a^p - b^q=1. Fortunately, the Catalan conjecture was proved, by Mihailescu (readable blog about it), between the paper being set, and me attempting it a few years later. I’m being flippant. This is not a standard trope in solving these questions. For very obvious reasons. If it can be killed by direct reference to a known theorem, it won’t be set.

Anyway, those references (and more to follow) are to illuminate why I thought this question was not too hard. Indeed, I feel one can make substantial meta-progress in your head. The given information is interesting, but for the purpose of this question is just a black box. By subtracting the expression for m from the expression for 2m, we can derive an expression for the required sum. It’ll be a quartic in m, because the leading terms won’t cancel.

This leaves all three of the methods above very accessible. Unfortunately m=0 would be a square were it not excluded specifically, so a modular arithmetic approach is unlikely to work directly. Bounding between two quadratics is entirely plausible, as is factorising and comparing number theoretic properties of the factors. I thought the second one seemed more promising, but either way, having two potentially good ideas based only on recent BMO problems before even writing anything down is a good opening.

We do have to calculate the sum, and I make it \frac{1}{4}m^2(5m+3)(3m+1). Now I’m not so sure how to bound this between two quadratics, because the leading coefficient is 15/4, which is not the square of a rational. But the factor analysis approach is definitely on.

Let’s review this generally. Throughout, suppose m,n are positive integers.

Claim 1: if mn is a square, then m and n are squares too.

Claim 2: if mn is a square, then m=n.

Both of these claims are false. However, a version of Claim 1 is true.

Claim 1′: if mn is a square, and m,n are coprime, then each is a square.

Even though this isn’t a named theorem, it is true, and well-known and can be used without proof. One way to prove it is to write m,n as products of primes, and show that since the primes are disjoint, the exponents must all be even. Most other methods will be equivalent to this, maybe with less notation.

What is good about Claim 1′ is that more complicated versions are true for for essentially similar reasons. For example

Claim 3: if mn is 6k^2, and m,n are coprime, then either one is a square and the other is six times a square; or one is two times a square, and the other is three times a square.

Claim 4: if mn is a square, and the greatest common divisor (m,n) is either 5 or 1, then either each is a square, or each is five times a square.

I cited some examples of the other methods I proposed. Here are some examples of this sort of thing in recent BMOs:

  • Q4 of BMO2 2016. Even the statement is suggestive. There are more complicated routes, but showing that (2p-u-v)(2p+u+v) is a square is one way to proceed, and then Claim 4 directly applies after checking a gcd.
  • Q2 of BMO1 2014 is similar, but it is much more explicit that this is the correct approach. Expose p^2 then use a (correct) version of Claim 2.
  • Q1 of BMO2 2009. Show that a and b must each be a square times 41 for rationality reasons.
  • Q6 of BMO1 2006. After sensible focused substitutions, obtain 3n^2=q(q-1). Rather than try to ‘solve’ this, extract the key properties along the lines of Claim 3, eliminate one of the cases by modular arithmetic, and return to the required statement.
  • Q3 of BMO2 2010 requires the student to reproduce the essentials of the arguments above in the case of a particular degree six polynomial with a tractable factorisation, along with some mild square-sandwiching or bounding arguments as discussed earlier.

In conclusion, I’m trying to say that if I claim I am confident I can find all integers m such that \frac14 m^2(5m+3)(3m+1) is a square, this is not based on complicated adult experience, but rather on recent problems at a similar sensible level. And I still don’t think it counts as Olympiad technique #371 – thinking about divisibility of factors is a good thing to do when talking about integers, and so it’s just a natural entry point into problems about squares. Plenty of problems might have this sort of thing as a starting point or an ending point.

For this problem we need a different ending point. To be brief, the factors (5m+3) and (3m+1) cannot both be squares because 5m+3 is never a square. So since the gcd of these factors is 1, 2 or 4, the only other option is that they are both squares times 2. And because -1 is not a square modulo 3, so 1 is not a (square times 2) modulo 3, and we are done. Note that this was a literal example of the first technique for proving something is not a square, proposed all the way back at the start of this section.

Footnotes

[3a] – some common themes for proving that sequences do include squares might be comparison with Pell’s Equations, or comparison with the explicit construction of solutions to Pythagoras’s equation.

Question Four

An example of an absorbing function is f(x)=\lfloor x\rfloor. One challenge is thinking of many other examples. This one is fine, but it’s true under replacing 2018 by 1 in the statement, and so it doesn’t really capture the richness of the situation.

Notation: the pre-image of a function is the language used to describe the inverse of a function which doesn’t have a uniquely-defined inverse. That is, if f is not injective, and multiple arguments have the output. We write f^{-1}(y)=\{x: f(x)=y\}. In particular, this is a set of values, not necessarily a single value. We also use \mathbb{Z} to denote the integers. We can apply pre-images to sets as well. So for example f^{-1}(\mathbb{Z})=\{x : f(x)\in \mathbb{Z}\}.

This question is tricky, and I will be surprised to see many full solutions from the eligible candidates. It rewards the sort of organisation and clear-thinking that is easier said than done in a time-pressured contest environment. There are also many many possible things to consider, and so is particularly challenging in the short timeframe of BMO2 as opposed to, for example, appearing as the middle question on a 4.5 hour international-level paper.

At a meta-level we are being asked to confirm or deny the existence of absorbing functions where f^{-1}(\mathbb{Z}) is small in some sense, firstly when actually having finite size, secondly when, although infinite, being a small sort of infinite, namely spread out in a sparse, well-ordered way (you might say countable if familiar with that language). The general idea is presumably that it’s hard to be absorbing if the pre-image of the integers is small, and so it’s reasonable to assume that it’s too hard if this is finite; but perhaps not quite too hard if it’s merely countable. So (no, yes) is a sensible guess at the answer to the question, though (no, no) might also fit, maybe with a harder argument for the second no.

Ok, instead of trying a) or b), just play with the configuration. Let A=f^{-1}(\mathbb{Z}). We will use this frequently. In the picture below, f maps the real line on top to the real line below. If two reals get mapped to the same image, then whether or not the image is an integer, the whole (closed) interval bounded by the two reals also gets mapped to the same image. This is because f is weakly increasing.

This means that A consists of various intervals (which include single points). But in both a) and b) we know that A is ‘small’, and so it cannot contain any intervals of positive length. So in fact A is a set of separated real values. In the case of a) it’s a finite set.

Do we want to try and iterate this, and look at f^{-1}(A)? Well maybe, but we don’t know much about about pre-images of A, only about pre-images of \mathbb{Z}.

But note that the pre-image of the pre-image of the … of the pre-image [2017 times] of A must be the whole real line, so at some point, some value has a pre-image that is an interval. So if we’re guessing that the answer to b) is yes, then we need to give a construction.

\mathbb{R} \stackrel{f}\longrightarrow ?? \stackrel{f}\longrightarrow\quad\ldots\quad \stackrel{f}\longrightarrow ??\stackrel{f}\longrightarrow A \stackrel{f}\longrightarrow f(A)\subset \mathbb{Z}.

If you play around for a bit, it seems very unlikely to be absorbing if the integers don’t get mapped to the integers. You can try to prove this, but at the moment we’re just aiming for a construction, so let’s assume f(\mathbb{Z})\subset \mathbb{Z}. It would be convenient if f(n)=n for all n\in \mathbb{Z}, but we already know that this won’t work because then the pre-image of the pre-image of the… of \mathbb{Z} is always \mathbb{Z}, but we need it to be \mathbb{R}.

The ideal situation would be if A= \mathbb{Z}\cup \{\ldots, a'_{-1},a'_0,a'_1,\ldots\}, where the pre-image of \{\ldots, a'_{-1},a_0,a'_1,\ldots\} is pretty much everything.

Informally, we are specifically banned from mapping intervals directly onto an integer. So have an intermediate set, and try to map almost everything (except the integers and the set itself) onto that set, so and map that set into the integers.

At this point, you really just have to have the right idea and finish it. Many things will work, but this seems the easiest to me. Let the set A consist of the integers and the (integers plus 1/2). And for x\in A, f(x)=2x. This is what f looks like so far.

Here the black crosses are integers, and the purple crosses are (integers plus 1/2). But now we need to make as many reals as possible in the top row map to a purple cross (which is allowed, because purple crosses aren’t integers), but we need also to preserve the weakly increasing property. Fortunately, we can exactly do that. Each cross of either colour in the top row maps to a black cross in the middle row (ie an integer), so we can map the open interval between crosses in the top row to a purple cross in the middle row. As shown in red:

Note that this is consistent. The fact that I haven’t drawn in the red cones into the bottom row is only because I didn’t use the bottom row to motivate doing this. I’ve shown a consistent definition of f that maps all the reals onto the integers in two steps. If it’s an integer to begin with, that was great; if it was an (integer plus 1/2) to begin with then it becomes an integer in one step and stays an integer; and otherwise it first maps to an (integer plus 1/2), and then to an integer in the second step.

To check you’ve understood, try to write down a standalone definition of this function.

I’ve therefore solved part b) with the alternative condition \ldots a_{-1}<a_0<a_1<a_2<\ldots which isn’t exactly as required. It requires one small and simple idea to convert to a solution to the actual statement. See if you can find it yourself!

I think part a) is harder, not because the solution will look more complicated, but because there are so many potential partial results you could try to prove, because there are so many sets you could consider. To name a few: the image of f, the image of f intersected with \mathbb{Z}, the image of \mathbb{Z}, the 2018-composition image f^{2018}(\mathbb{R}), the 2018-composition image f^{2018}(\mathbb{Z}) and so on and so forth. You might have good insight into the wrong things.

For me, the crucial observation (which you can see from the figure in the b) construction) is that when composing an increasing function with itself, the ‘trajectories’ are either increasing or decreasing. That is, if x\le f(x) (respectively, x\ge f(x)), then x\le f(x)\le f^2(x)\le f^3(x)\le\ldots (respectively x\ge f(x)\ge f^2(x)\ge \ldots). Again, you can think of this as Olympiad technique #371 if you insist, but I don’t think that’s helpful. There are lots of things one could try to say here, and this turns out to be natural, true and useful, but you can’t know it’s useful until you play with it.

Anyway, we’re playing with part a), and we know that f^k(x) is an integer for all large enough k, and that f^{k+1}(x) is also an integer, so f^k(x) is one of a finite set of integers because of the condition on A. But we’ve seen the sequence x,f(x),f^2(x),\ldots is weakly increasing or weakly decreasing, and so if we also know it’s eventually bounded (because eventually it’s in this finite set) then it must eventually be constant. And this constant is one of the integers, say n. But unless we started from n, this means that f(n)=n, but also f(x)=n for some other real value x. And so exactly as at the very very beginning, that’s bad, because then the whole interval [x,n] gets mapped to n, which is a contradiction.

Question Two – Origin story

The origin story for Q2 started in a talk I heard by Renan Gross at Weizmann, who referenced some of the history of Scenery Reconstruction. Roughly speaking, we colour the integers (say with two colours), and then let loose a random walker, who tells us the sequence of colours she observes during her walk, but no other information about the walk itself.

How much information can we recover about the colouring? Obviously, the best we can hope for is to recover the colouring, up to translations and reflection, since for every possible random walk trajectory, the exact reflection is equally probable, and we are given no information about the starting point.

Since lots of the transitions between recoverable and unrecoverable depend on the periodicity of the colouring, a reasonable toy model is to do it on a cycle. Note that the Strong Law of Large Numbers tells us that we almost surely recover the number of black sites and white sites from an the infinite trajectory of the random walk. Of course it’s possible that there are only two black vertices, and they are adjacent, and the walker oscillates between them, thus seeing BBBBBB… But this is extremely unlikely. You could think of this in Bayesian terms as strongly increasing the prior on the whole cycle being black, but I think initially it’s best to do this as an infinite-time, SLLN problem not as finite time WLLN/CLT reweightings of anything.

But what more? It’s clear that the lengths of all black substrings should follow some mixed geometric-ish distribution, and this distribution will almost surely wash out as the empirical distribution in an SLLN sense. But it’s tricky to justify why such a mixed geometric-ish distribution should be uniquely determined by the lengths of black arcs in the cycle. But it does definitely feel like we should have enough information to reconstruct the colouring up to reflection/rotation with probability one. For example, analogously to the number of black vertices and the number of white vertices, we should be able to recover the number of adjacent black vertices, the number of adjacent white vertices, and the number of black-white adjacent vertices, and so on.

Anyway, this can be done, and it follows as a consequence of various authors’ work answering some more general conjectures of Benjamini and, separately, of den Hollander and Keane. Douglas Howard [DH] shows a handful of generalisations of this, as do Benjamini and Kesten [BK]. Most of this work is focused on sceneries on \mathbb{Z}, but periodic sceneries are often used as a basis, and of course, the only difference between periodic sceneries on \mathbb{Z} and sceneries on the N-cycle are whether you know the period in advance. [BK] show that ‘almost all’ sceneries are distinguishable in a particular sense, in response to which Lindenstrauss [L99] exhibits a large family of sceneries which are not distinguishable. A readable but technical review is [ML].

So Renan’s talk was about the similar problem (and generalisations) on the hypercube [GG]. Rather than paraphrase the main differences badly, you can read his own excellent blog post about the work.

On the train back to Haifa from Rehovot, I was thinking a bit about the cycle case, and what happens if you generalise the random walk with varying jump lengths, or indeed introduce a demon walker, whose goal is to make it as hard as possible for the reviewer to deduce the colouring. One way this can certainly happen is if the walker can avoid visiting some particular site, as then how could one possibly deduce the colour of the never-visited site? And so we get to the statement posed.

References

[BK] – Benjamini, Kesten, 1996 – Distinguishing sceneries by observing the scenery along a random walk path

[dH] – den Hollander, 1988 – Mixing properties for random walk in random scenery

[DH] – Douglas Howard, 1996 – Detecting defects in periodic scenery by random walks on Z

[GG] – Grupel, Gross, 2017 – Indistinguishable sceneries on the Boolean hypercube

[L99] – Lindenstrauss, 1999 – Indistinguishable sceneries

[ML] – Matzinger, Lember, 2003 – Scenery reconstruction: an overview [link]

 

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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.