Egyptian Fractions – 4
Around 4000 years ago, the ancient Egyptians developed a system where all their fractions were of the form
1/n. These are called Egyptian Fractions. Some Egyptian Fractions can be written as the difference of two
Egyptian Fractions. ⅓ = ½ – ⅙ is an example of doing that.
THE CHALLENGE
Which Egyptian Fractions can be written as a difference of two Egyptian Fractions in only
one way? Which ones can be written in exactly two ways? Are there some that have more than two ways?
1/N = 1/A – 1/B
Notes
THE CHALLENGE
One thing to note is that for any n, there are a limited number of possible solutions. If 1/N = 1/A – 1/B, then 1/A – 1/N = 1/B > 0. 1/A – 1/N > 0 means that N > A. So, 1 < A < N, which means there are at most N – 2 solutions, and they can be found, somewhat tediously, by checking all the values of A in that range.
For denominators that are even numbers, say 2d, there is always 1/2d = 1/d – 1/2d.
Another way to produce differences is to start with an addition formula. The simplest one we have is that 1/n = 1/(n+1) + 1/[n(n+1)]. This becomes 1/(n+1) = 1/n – 1/[n(n+1)]. A more general version of this is 1/ab = 1/[a(a+b)] + 1/[b(a+b)] which becomes 1/[a(a+b)] = 1/ab – 1/[b(a+b)] where ab > 1.
So, whenever the denominator can be factored as a(a+b), with a and b positive and ab > 1, we will have a difference. In particular, because every number n > 2 can be written as n = 1 x (1 + (n-1)), then every Egyptian Fraction can be written as a difference in at least one way. Here are a few examples:
- ⅓: a = 1, b = 2 gives ½ – ⅙
- ¼: a = 1, b = 3 gives ⅓ – 1/12
- ⅕: a = 1, b = 4 gives ¼ – 1/20
- ⅙: a = 1, b = 5 gives ⅕ – 1/30; a = 2, b = 1 gives ½ – ⅓
- 1/7: a = 1, b = 6 gives ⅙ – 1/42
- ⅛: a = 1, b = 7 gives 1/7 – 1/56; a = 2, b = 2 gives ¼ – 1/8
This is not everything. For example, we have missed ⅙ = ¼ – 1/12.
Diving into some algebra, for a > 1 and b > a >1, consider 1/n = 1/a – 1/b = (b – a) / ab. Clearing the denominators produces ab = n(b-a), and solving for n gives, n = ab/(b-a). Whenever we can find a and b that makes this work, we will be able to produce the difference.
Let’s look at the complete solutions for a few examples from this point of view and see if we can spot a pattern.
- 3 = 2 x 6 / (6 – 2)
- 4 = 2 x 4 / (4 -2) = 3 x 12 / (12 – 3)
- 5 = 4 x 20 / (20 – 4)
- 6 = 2 x 3 / (3 – 2) = 3 x 6 / (6 – 3) = 4 x 12 / (12 – 4) = 5 x 30 / (30 – 5)
- 7 = 6 x 42 / (42 – 6)
- 8 = 4 x 8 / (8 – 4) = 6 x 24 / (24 – 6) = 7 x 56 / (56 – 7)
Looking at these examples, it seems likely that there is only one solution when n is a prime. Let’s see why that is. If n is a prime and a < n, then n = a x b / (b – a) forces b to be a multiple of n, say b = n x c. Then n = a x nc / (nc – a) says that ac / (nc – a) = 1, which means ac = nc – a. Rewriting this we get nc = a(c + 1). Because n is a prime, we have c + 1 = n and a = c = n – 1. So a = n – 1 and b = n(n-1) is the only solution