T-shirt from Egypt
- What on earth are you wearing?
- It is a T-shirt I bought in Cairo on my last holiday.
- T-shirt with math on it! Only you would be crazy enough to pay for something like that.
- I didn’t. I got it for free.
- How come?
- It was free for those who could prove that the equation has no solution when the xs are odd integers.
- And that you call a holiday?!
- Actually, it gets better.
- Do explain.
- I got another one for free as well.
- Of the same kind?
- I proved that there is a solution when the xs are distinct positive integers.
- Do you mind if I open the window?
Problem source: Math problems from PIMS.
September 28th, 2009 at 2:01 am
1/3 + 1/6 = 1/2 ___ [1]
=> 1/3n + 1/6n = 1/2n ___ [2]
1/2 + 1/3 + 1/6 = 1 ___ [3]
=> 1/2n + 1/3n + 1/6n = 1/n ___ [4]
so, to get such an equation with any desired number of terms on the LHS, start with equation [1] if you want an even number of terms and equation [3] if you want an odd number of terms, then repeatedly replace the smallest term on the LHS (i.e. the one whose denominator is largest) by 3 terms as per equation [4]. Equation [2] offers an alternative way to increase the number of terms by 1 if you have a term with an even denominator.
I say the smallest term to ensure that no new terms equal any of the unaffected terms (so that the terms are still all distinct).
Is it possible with all the denominators odd? That seems like finding out whether or not there is an odd perfect number. There is no obvious way to see that it isn’t possible, but no obvious solution either.
September 28th, 2009 at 2:20 am
Come to think (again!) of it, in the special case presented here, there is no solution. More generally:
Theorem. There are no solutions to
1/x_1 + … + 1/x_k = 1/x_[k+1]
if all the x_i are odd integers, and k is even.
Proof:
Let L be the lowest common denominator of the x_i. Then L is odd, and, for each i,
1/x_i = n_i/L
where n_i is an integer (by definition of L) and is odd, as all factors of an odd integer are odd. Then
(n_1 + … + n_k)/L = n_[k+1]/L
=> n_1 + … + n_k = n_[k+1]
The RHS is odd. The LHS is the sum of an even number of odd terms, and is thus even. Contradiction. So there are no solutions.
This still doesn’t settle the more general question; it merely shows that the number of terms on the LHS must be odd.