My book on abstract algebra had a small puzzle in it I was interested in. It asks the reader to prove that
$\sum\limits_{j=1}^{n}\frac{1}{x_j} = 1$ only has finitely many solutions where $x_j \in \mathbb{N}$.
My first attempt to proof it was with induction. It's easy to see that for $n=1$ it only has $1$ solution. For the induction step I assume that in the case $n=k+1$ we have infinitely many solutions, so it follows that $\sum\limits_{j=1}^{k}\frac{1}{x_j} = \frac{x_k-1}{x_k}$ which implies that $\sum\limits_{j=1}^{k}\frac{1}{x_j} = \frac{1}{x_k}$ (assuming $x_k \neq 1$).
This is pretty much as far as I got. I'd really love to have a hint rather than a full solution if possible.
Hint: First order the $x_i$ according to size. Then try all possibilities for $x_1$, then $x_2$, etc. The constraints that $x_i\ge x_{i-1}$ and that $\frac{n-i+1}{x_i}\ge 1-\sum_{j=1}^{i-1}x_j$ mean that one only has to try finitely many possibilities for $x_i$ at each step.