# Is it possible to write 1 as the sum of the reciprocals of x odd integers? x is the odd number. [duplicate]

Actually, thats it. The question is: Is it possible to write 1 as the sum of the reciprocals of x odd integers? x is the odd number. For x - even the impossibility is obvious. But what to do for x - odd?

I have seen some of the similar questions, but almost everywhere there were odd numbers in a row (2n+1,n=1,2...) And the answer was using the Harmonic number. Somewhere it is mentioned that the difference of harmonic numbers can't be an integer, but i don't know if it is true... If it is, then it is possible to write this (finite) series as a combination of harmonic numbers.

Please, correct me if i'm wrong.

Image

## marked as duplicate by Jyrki LahtonenApr 3 '18 at 12:56

• Like $1/3 +1/3+1/3$? – Randall Sep 24 '17 at 17:51
• $1=\frac 11$ seems to work, but you may want to exclude the trivial case – Mark Bennet Sep 24 '17 at 17:54
• From the article on Egyptian Fractions in wikipedia "It is known that every x/y with odd y has an expansion into distinct odd unit fractions, constructed using a different method than the greedy algorithm." en.wikipedia.org/wiki/Egyptian_fraction. The article gives some references which may give more detail of the general theory. – Mark Bennet Sep 24 '17 at 18:23
• You: For x - even the impossibility is obvious What does it mean? There exists the expansion $$1=\frac12 + \frac14 + \frac16 + \frac{1}{12}$$ where all denominators are even and the number of fractions is even as well. There are many other examples in the link given by Professor Vector in a comment to an answer below, that is Egyptian Fractions. – Jeppe Stig Nielsen Sep 24 '17 at 21:46
• Sorry, I see what you meant. If $1$ is written as a sum of reciprocals of odd numbers, then the number of terms must be odd. If not, you could multiply each side of the equation by the product of all the denominators to get an odd integer written as the sum of an even number of odd addends, a contradiction since modulo $2$ we have $1+1+1+\dots+1=(1+1)+(1+1)+\dots+(1+1)\equiv 0+0+\dots+0=0$. – Jeppe Stig Nielsen Sep 24 '17 at 22:08

Also sprach Mathematica: $$1-1/3 - 1/5 - 1/7 - 1/9 - 1/11 - 1/15 - 1/21 - 1/135 - 1/10395=0.$$

• Undoubtedly there are many solutions. I was simply filling it up with reciprocals of small odd integers, and then experimented a bit when I got close to zero. – Jyrki Lahtonen Sep 24 '17 at 18:25
• That is great! I'm now interested in the solution without prime numbers... but i don't think it would be good to ask almost the same question. Thank you, this will help. – Andrew Sep 24 '17 at 18:27
• 9 is the smallest possible number of fractions needed, but there are several solutions, look here. – Professor Vector Sep 24 '17 at 19:18

Here's a technique to generate many solutions:

Take an odd abundant number ($945$ is the smallest) and (try to) find a subset of its proper factors that adds to the number:

$7+9 + 15+27+35+45+63+105+135+189+315=945$

Divide by the original number

$\frac{7+9 + 15+27+35+45+63+105+135+189+315}{945}=1$

Write the left hand side as a sum of simplified fractions:

$\frac{1}{135}+\frac{1}{105}+\frac{1}{63}+\frac{1}{35}+\frac{1}{27}+\frac{1}{21}+\frac{1}{15}+\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}=1$

• This is a great insight – 6005 Sep 25 '17 at 2:22
• According to Wikipedia weird number it is an open problem whether any odd abundant number will work (the "try to" parenthesis above). So if you pick an odd number where the divisors are plenty enough that the sum of all of them is too great, can we pick a subset of the divisors that sum to the odd number itself? – Jeppe Stig Nielsen Sep 25 '17 at 7:16

It's possible. I can show to you that it's possible and a constraint for generating such a sequence.

We have $k_m =$ any integer, $O_m =$ any odd number where each $O_m$ is distinct and $E_m =$ any even number. We can express it thusly:

$$\frac{k_1}{O_1} + \frac{k_2}{O_2} + \frac{k_3}{O_3} + \ ... +\frac{k_n}{O_n} = 1$$

Note that $n$ is odd.

Now multiply everything by $O_1$: $$k_1 + O_1\frac{k_2}{O_2} + O_1\frac{k_3}{O_3} + \ ... +O_1\frac{k_n}{O_n} = O_1$$

By $O_2$:

$$O_2k_1 + O_1k_2 + O_2O_1\frac{k_3}{O_3} + \ ... + \ O_2O_1\frac{k_n}{O_n} = O_1O_2$$

Continue doing this for every $O$, this will yield the following: $$(\overbrace{O_2O_3O_4...}^{n-1})k_1 + (\overbrace{O_1O_3O_4...}^{n-1})k_2 + \ ... + \ (\overbrace{O_2O_3O_4...}^{n-1})k_n = O_2O_3O_4...O_n$$

So what happens here is that every $k_m$ gets multiplied by every $O$ except $O_m$ (because $O_m$ is already in the denominator so they cancel each other out). This will then equal the product of every $O$ times $1$ on the right hand side.

We know that $n$ is odd so $n-1$ must be even, therefore we have an even number off odd numbers multiplied together which is an odd number. This means we can rewrite the equation thusly (odd = odd):

$$k_1O + k_2O + k_3O + \ ... \ + k_nO = O_1O_2O_3...O_n$$

The right hand side $O_1O_2O_3...O_n$ is an odd number of odd integers multiplied together, which is just an odd number. So if we even want a chance for them to be the same, the left hand side needs to be odd as well. We have an odd number $n$ of terms there, and each $k$ multiplied by an odd number $O$. So there needs to be an odd number of $k$ that is odd, and an even number of $k$ that is even. This is the only way that we can get an odd number on the left hand side.

So it's possible because we don't arrive at a contradiction by setting the sum to equal $1$. I have no idea about a specific sequence like this though, but every sequence must follow the pattern I mentioned.

EDIT First had a mistake, corrected it.

• The product of odd numbers is odd – Andrew Sep 24 '17 at 18:34
• You can't go from $O_1 O_2 \dots$ to an even number. – qwr Sep 24 '17 at 18:46
• I agree, made a mistake earlier. – Sam Anderson Sep 24 '17 at 19:16