1 = 1/a + 1/b + ...

We will prove that 1 can never be written as an even number of odd number reciprocals.

Let the odd numbers be $a_1, a_2, \ldots , a_{2n}$ . Suppose we have

$1 = \frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_{2n}}$

Now, let $b_k = a_1a_2 \ldots a_{2n}/a_k$ (the product of all $a_i$ except $a_k$ ). Multiplying both sides with $a_1a_2 \ldots a_{2n}$ , we get

$a_1a_2 \ldots a_{2n} = b_1 + b_2 + \ldots + b_{2n}$

The product of odd numbers is always odd, so $a_1a_2 \ldots a_{2n}$ and $b_k$ has to be odd. The sum of $2n$ odd numbers (in this case $b_k$ ), however, is always even for any whole number $n$ .

Since both sides have different parities (LHS is odd, RHS is even), the Diophantine equation has no solutions.

Of course, it is then an immediate result that 10 odd number reciprocals will never sum to 1.

its really easy. we say $1=\sum_{n=1}^{10}\dfrac{1}{2a_n-1}$ where $a_n$ is a integer. now we multiply: $\prod_{n=1}^{10} (2a_n-1)=\sum_{0<b_1<b_2<...<b_9<11} (\prod_{n=1}^{9} (2a_{b_n}-1))$ we see that LHS is odd as it is multiple of odds only. the product on the RHS becomes odd. but when we add 10 odds, it becomes even. since LHS=odd,RHS=even, they cannot be equal. genarally 1 can not be written as sum of reciprocals of 2n odd integers.