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Worked problem 2 () 1999 Paper I

The n positive numbers x1,x2,,xn, where n 3, satisfy

x1 = 1 + 1 x2,x2 = 1 + 1 x3,,xn1 = 1 + 1 xn,

and also

xn = 1 + 1 x1.

Show that

x1,x2,,xn > 1,
x1 x2 = x2 x3 x2x3 ,
x1 = x2 = = xn.

Hence find the value of x1.

First thoughts

My first thought is that this question has an unknown number of variables: x1, … , xn. That makes it seem rather complicated. I might, if necessary, try to understand the result by choosing an easy value for n (maybe n = 3). If I manage to prove some of the results in this special case, I will certainly go back to the general case: doing the special case might help me tackle the general case, but I don’t expect to get many marks in an exam if I just prove the result in one special case.

Next, I see that the question has three sub-parts, then a final one. The final one begins ‘Hence ...’. This means that I must use at least one of the previous parts in my working for the final part. It is not clear from the structure of the question whether the three sub-parts are independent; the proof of (ii) and (iii) may require the previous result(s), or it may not.

Actually, I think I can see how to do the very last part. If I assume that (iii) holds, so that x1 = x2 = = xn = x (say), then each of the equations given in the question is identical and each gives a simple equation for x.

It is surprising, isn’t it, that the final result doesn’t depend on the value of n? That makes the idea of choosing n = 3, just to see what is going on, quite attractive, but I’m not going to resort to that idea unless a get very stuck.

The notation in part (i) is a bit odd. I’m not sure that I have seen anything like it before.13 But it can only mean that each of the variables x1, x2, xn is greater than 1.

In fact, now I think about it, I am puzzled about part (i). How can an inequality help to derive the equalities in the later parts? I can think of a couple of ways in which the result xi > 1 could be used. One is that I may need to cancel, say, x1 from both sides of an equation in which case I would need to know that x10. But looking back at the question, I see I already know that x1 > 0 (it is given right at the start of the question), so this cannot be the right answer. Maybe I need to cancel some other factor, such as (x1 1). Another possibility is that I get two or more solutions by putting x1 = x2 = = xn = x and I need the one with x > 1. This may be the answer: looking back at the question again, I see that it asks for the value of x1 — so I am looking for a single value. I’m still puzzled, but I will remember to keep a sharp look out for ways of using part (i).

One more thing strikes me about the question. The equations satisfied by the xi are given on two lines (‘and also’). This could be for typographic reasons (the equations would not all fit on one line) but more likely it is to make sure that I have noticed that the last equation is a bit different: all the other equations relate xi to xi+1, whereas the last equation relates xn to x1. It goes back to the beginning, completing the cycle. I’m pleased that I thought of this, because this circularity must be important.

Doing the question

I think I will do the very last part first, and see what happens.

Suppose, assuming the result of part (iii), that x1 = x2 = = xn = x. Then substituting into any of the equations given in the question gives

x = 1 + 1 x

i.e. x2 x 1 = 0. Using the quadratic formula gives

x = 1 ±5 2 ,

which does indeed give two answers (despite the fact that the question asks for just one). However, I see that one is negative and can therefore be eliminated by the condition xi > 0 which was given in the question (not, I note, the condition x1 > 1 from part (i); I still have to find a use for this).

I needed only part (iii) to find x1, so I expect that either I need both (i) and (ii) directly to prove (iii), or I need (i) to prove (ii), and (ii) to prove (iii).

Now that I have remembered that xi > 0 for each i, I see that part (i) is obvious. Since x2 > 0 then 1x2 > 0 and the first equation given in the question, x1 = 1 + 1x2, shows immediately that x1 > 1 and the same applies to x2, x3, etc.

Now what about part (ii)? The given equation involves x1, x2 and x3, so clearly I must use the first two equations given in the question:

x1 = 1 + 1 x2,x2 = 1 + 1 x3.

Since I want x1 x2, I will see what happens if I subtract the two equations:

x1 x2 = (1 + 1 x2) (1 + 1 x3) = 1 x2 1 x3i = x3 x2 x2x3 = x2 x3 x2x3 . ()

That seems to work!

One idea that I haven’t used so far is what I earlier called the circularity of the equations: the way that xn links back to x1. I’ll see what happens if I extend the above result. Since there is nothing special about x1 and x2, the same result must hold if I add 1 to each of the suffices:

x2 x3 = x3 x4 x3x4 .

I see that I can combine this with the previous result:

x1 x2 = x2 x3 x2x3 = x3 x4 x2x32x4.

I now see where this is going. The above step can be repeated to give

x1 x2 = x3 x4 x2x32x4 = x4 x5 x2x32x42x5 =

and eventually I will get back to x1 x2:

x1 x2 = x4 x5 x2x32x42x5 = = ± x1 x2 x2x32x42x52xn2x12x2


(x1 x2) 1 1 x12x22x32x42x52xn2 = 0.

I have put in a ± because each step introduces a minus sign and I’m not sure yet whether the final sign should be (1)n or (1)n1. I can check this later (for example, by working out one simple case such as n = 3); but I may not need to.

I deduce from this last equation that either

x12x 22x 32x 42x 52x n2 = ±1


x1 = x2

(which is what I want). At last I see where to use part (i): I know that

x12x 22x 32x 42x 52x n2 ± 1

because x1 > 1, x2 > 1, etc. Thus the only possibility is x1 = x2. Since there was nothing special about x1 and x2, I deduce further that x2 = x3, and so on, as required.


There were a number of useful points in this question.

The first point concerns using the information given in the question. The process of teasing information from what is given is fundamental to the whole of mathematics. It is very important to study what is given (especially seemingly unimportant conditions, such as xi > 0) to see why they have been given. If you find you reach the end of a question without apparently using some given information, then you should look back over your work: it is very unlikely that a condition has been given that is not used in some way. It may not be a necessary condition — and we will see that the condition xi > 0 is not, in a sense, necessary in this question — but it should be sufficient. The other piece of information in the question which you might easily have overlooked is the use of the singular rather than the plural in referring to the solution (‘ ... find the value of ...’), implying that there is just one value, despite the fact that the final equation is quadratic.
The second point concerns the structure of the question. Here, the position of the word ‘Hence’ suggested strongly that none of the separate parts were stand-alone results; each had to be used for a later proof. Understanding this point made the question much easier, because I was always on the look out for an opportunity to use the earlier parts. Of course, in some problems (without that ‘hence’) some parts may be stand-alone; though this is rare in STEP questions.

You may think that this is like playing a game according to hidden (STEP) rules, but that is not the case. Precision writing and precision reading is vital in mathematics and in many professions (law, for example). Mathematicians have to be good at it, which is the reason why so many employers want to recruit people with mathematical training.

The third point was the rather inconclusive speculation about the way inequalities might help to derive an equality. It turned out that what was actually required was x1x2xn ± 1. I was a bit puzzled by this possibility in my first thoughts, because it seemed that the result ought to hold under conditions different from those given; for example, xi < 0 for all i (does this condition work??). Come to think of it, why are conditions given on all xi when they are all related by the given equations? This makes me think that there ought to be a better way of proving the result which would reveal exactly the conditions under which it holds.
Then there was the idea (which I didn’t actually use) that I might try to prove the result for, say, n = 3 to help me understand what was going on. This would not have counted as a proof of the result (or anything like it), but it might have given me ideas for tackling the question.
A key observation was that the equations given in the question are ‘circular’. It was clear that the circularity was essential to the question and it turned out to be the key to the most difficult part. Having identified it early on, I was ready to use it when the opportunity arose.
Finally, I was pleased that I read the whole question carefully before plunging in. This allowed me to see that I could easily do the last part before the preceding parts, which I found very helpful in getting into the question.

Final thoughts

It occurs to me only now, after my post-mortem, that there is another way of obtaining the final result.

Suppose I start with the idea of circularity (as indeed I might have, had I not been otherwise directed by the question) and use the given equations to find x1 in terms of first x2, then x3, then x4 and eventually in terms of x1 itself. That should give me an equation I can solve, and I should be able to find out what conditions are needed on the xi. Try it. You may need to guess a formula for x1 in terms of xi from a few special cases, then prove it by induction.14 You will find it useful to define a sequence of numbers Fi such that F0 = F1 = 1 and Fn+1 = Fn + Fn1. (These numbers are called Fibonacci numbers.15)

You should find that if xn = x1 for any n (greater than 1), then xn = 1±5 2 .

13 It is just the sort of thing that is used in university texts; but I’m not sure that it would be used in STEP papers nowadays.

14 I went as far as x7 to be sure of the pattern: I found that x7 = 8x1+5 5x1+3.

15 Fibonacci (short for filius Bonacci — son of Bonacci) was called the greatest European mathematician of the Middle Ages. He was born in Pisa (Italy) in about 1175 AD. He introduced the series of numbers named after him in his book of 1202 called Liber Abbaci (Book of the Abacus). It was the solution to the problem of the number of pairs of rabbits produced by an initial pair: A pair of rabbits are put in a field and, if rabbits take a month to become mature and then produce a new pair every month after that, how many pairs will there be in twelve months time?