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Problem 11:  Egyptian fractions () 2000 Paper II

A number of the form 1 N , where N is an integer greater than 1, is called a unit fraction.

Noting that

1 2 = 1 3 + 1 6 and  1 3 = 1 4 + 1 12,

guess a general result of the form

1 N = 1 a + 1 b ()

and hence prove that any unit fraction can be expressed as the sum of two distinct unit fractions.

By writing () in the form

(a N)(b N) = N2

and by considering the factors of N2, show that if N is prime, then there is only one way of expressing 1 N as the sum of two distinct unit fractions.

Prove similarly that any fraction of the form 2 N , where N is prime number greater than 2, can be expressed uniquely as the sum of two distinct unit fractions.


Fractions written as the sum of unit fractions are called Egyptian fractions: they were used by Egyptians. The earliest record of such use is 1900BC. The Rhind papyrus in the British Museum gives a table of representations of fractions of the form 2 n as sums of unit fractions for all odd integers n between 5 and 101 — a remarkable achievement when you consider that algebra was 3,500 years in the future.

It is not clear why Egyptians represented fractions this way; maybe it just seemed a good idea at the time. Certainly the notation they used, in which for example 1 n was denoted by n with an oval on top, does not lend itself to generalisation to fractions that are not unit. One of the rules for expressing non-unit fractions in terms of unit fractions was that all the unit fractions in should be distinct, so 2 7 had to be expressed as 1 4 + 1 28 instead of as 1 7 + 1 7, which seems pretty daft.

It is not obvious that every fraction can be express in Egyptian form; this was proved by Fibonacci in 1202. There are however still many unsolved problems relating to Egyptian fractions.

Egyptian fractions have been called a wrong turn in the history of mathematics; if so, it was a wrong turn that favoured style over utility; no bad thing, in my opinion.

Solution to problem 11

A good guess would be that the first term of the decomposition of 1 N is 1 N + 1, i.e. a = N + 1. In that case, the other term is 1 N 1 N + 1 i.e. b = N(N + 1). That proves the result that every unit fraction can be expressed as the sum of two unit fractions.

The only factors of N2 (since N is prime) are 1, N and N2. The possible factorisations of N2 are therefore N2 = 1 × N2 or N2 = N × N. We discount N × N, since this would lead to a = b, and this is ruled out in the question. That leaves only a N = 1 and b N = N2 (or the other way round). Thus N + 1 and N2 + N are the only possible values for a and b and the decomposition is unique.

For the second half, set

2 N = 1 a + 1 b

where ab. We need to aim for an equation with N2 on one side, so that we can use the method of the first part. We have ab 1 2(a + b)N = 0 which we write as

a N 2 b N 2 = N2 4 i.e. 2a N 2b N = N2.

Thus 2a N = N2 and 2b N = 1 (or the other way round). The decomposition is therefore unique and given by

2 N = 1 1 2(N2 + N) + 1 1 2(N + 1).

The only possible fly in the ointment is the 1 2 in the denominators: a and b are supposed to be integers. However, all prime numbers greater than 2 are odd, so N + 1 and N2 + N are both even and the denominators are indeed integers.


Another way of getting the last part (less systematically) would have been to notice that

1 N = 1 (N + 1) + 1 N(N + 1) 2 N = 1 1 2(N + 1) + 1 1 2N(N + 1)

which gives the result immediately. How might you have noticed this? Well, I noticed it by trying to work out some examples, starting with what is given at the very beginning of the question. If 1 3 = 1 4 + 1 12 then 2 3 = 2 4 + 2 12 which works.

Of course, the advantage of the systematic approach is that it allows generalisation: what happens if N is odd but not prime; what happens if the numerator is 3 not 2?