For every $n\ge3$ there exists $n$ different integers $a_1,...,a_n$ so that $\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}=1$

I have a hard induction problem here that I feel completly lost on.

For every $n\ge3$ there is n different positive integer $a_1,...,a_n$ so that $\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}=1$.

There is only one possible solution for $n=3 $

$\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$

for $n=4$ there is 2 possible solutions and they both start with $\frac{1}{2}$.

$\frac{1}{2}+\frac{1}{3}+\frac{1}{8}+\frac{1}{24}$ and $\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{20}$.

I think the first one is the one.

For $n=5$ there is alot of different possible solutions but it have to start with at least $\frac 13$

I think it have to start with $\frac 12+\frac13$ and then be some kind of sum answer for the last $\frac16$ based on $n$. But I don't know

I have tried finding a pattern but not getting anywhere anymore. Any tips on how to tackle a problem like this?

Hint: using the identity $\displaystyle\;\frac{1}{k}=\frac{1}{k+1}+ \frac{1}{k(k+1)}\,$ and repeated halving:

• $\displaystyle\frac{1}{2}+\frac{1}{2}=1 \;\;\to\;\; \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$

• $\displaystyle\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1 \;\;\to\;\; \frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{20}=1$

• $\displaystyle \displaystyle\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=1 \;\;\to\;\; \ldots$

Induction step:

Well if $\sum_{i=1}^n \frac 1{k_i} = 1$ where $k_i > 1$ and $k_i$ are distinct, then

$\sum_{i=1}^n \frac 1{2k_i} = \frac 12$ where $2k_i > 2$ and $2k_i$ are distinct.

So Let $m_i = 2k_i$ and $\sum_{i=1}^n\frac 1{m_i} = \frac 12$ where $m_i > 2$ and $m_i$ are distinct.

So let $m_{n+1} = 2$ then $\sum_{i=1}^{n+1}\frac 1{m_i} = \frac 12 + \frac 12 = 1$ where $m_i > 1$ and $m_i$ are distinct.

Base step: $\frac 12 + \frac 16 + \frac 13 = 1$

(So $\sum\limits_{k=1}^{n-3} \frac 1{2^k} + \frac 1{2*2^{n-3}} +\frac 1{6*2^{n-3}} + \frac 1{3*2^{n-3}} =1$. Or $\sum\limits_{k=1}^{n-2} \frac 1{2^k} + \frac 1{3*2^{n-2}} + \frac 1{3*2^{n-3}}=1$. Which is sort of interesting.)

($\frac 12 + \frac 14 + ..... + \frac 1{2^{m}} = \frac {2^{m}-1}{2^{m}}$ and so

$\sum\limits_{k=1}^{n-2} \frac 1{2^k} + \frac 1{3*2^{n-2}} + \frac 1{3*2^{n-3}}=\frac {2^{n-2}-1}{2^{n-2}} + + \frac 1{3*2^{n-2}} + \frac 2{3*2^{n-2}}=$

$\frac {2^{n-2}-1}{2^{n-2}} + \frac 13*\frac 1{2^{n-2}}(2 + 1)= \frac {2^{n-2}-1}{2^{n-2}}+\frac 1{2^{n-2}} = \frac {2^{n-2}}{2^{n-2}} = 1$.)

For $n=3$: $1/2+1/3+1/6=1$

For $n=4$: $1/2+1/4+1/6+1/12=1$

Going from $n$ to $n+2$: Let's say the smallest fraction in the sum for $n$ is $1/m$. Replace $1/m$ with $$\frac{1}{m} \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \right) = \frac{1}{2m} + \frac{1}{3m} + \frac{1}{6m}$$