Arabian Nights!

I'd love to know if there's a non-case bashing approach to this, but as I didn't find one, here goes!

The first son gets $\frac{N+1}{u}$ camels; the second $\frac{N+1}{v}$ and the third $\frac{N+1}{w}$ . The total of these three amounts has to be $N$ , so that Calvin can ride off into the sunset on his own camel at the end. In other words, $\frac{N+1}{u}+\frac{N+1}{v}+\frac{N+1}{w}=N$

Dividing through by $N+1$ , $\frac{1}{u}+\frac{1}{v}+\frac{1}{w}=\frac{N}{N+1}$

Adding $\frac{1}{N+1}$ to both sides, $\frac{1}{u}+\frac{1}{v}+\frac{1}{w}+\frac{1}{N+1}=1$

Since we're after unordered triples $\{u,v,w\}$ , without loss of generality we may assume that $u \le v \le w$ . Also, since $w$ divides $N+1$ , we must have $w \le N+1$ .

This is enough to reduce the number of cases to something manageable. Using the inequalities above, $\frac{1}{u}<\frac{1}{u}+\frac{1}{v}+\frac{1}{w}+\frac{1}{N+1} \le \frac{4}{u}$ so $1<u \le 4$ .

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Case $u=2$ : The equation becomes $\frac{1}{v}+\frac{1}{w}+\frac{1}{N+1}=\frac12$

Using the same reasoning as above, we find $2<v \le 6$ . Carrying on in this way, we find $10$ solutions for $(u,v,w,N)$ : $\begin{matrix} (2,3,7,41) & (2,3,8,23) & (2,3,9,17) & {\color{#D61F06} (2,3,10,14)} & (2,3,12,11) \\ (2,4,5,19) & (2,4,6,11) & (2,4,8,7) & (2,5,5,9) & (2,6,6,5) \end{matrix}$

Of these solutions, one (highlighted in red) doesn't work; $N+1$ is not a common multiple of $u,v,w$ .

So there are $9$ valid solutions with $u=2$ .

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Case $u>2$

In the same way, we find the following additional solutions $\begin{matrix} (3,3,4,11) & (3,3,6,5) & {\color{#D61F06} (3,4,4,5)} & (4,4,4,3) \end{matrix}$

Again, one of these doesn't work; so there are a further $3$ valid solutions with $u>2$ .

In total, there are $\boxed{12}$ valid solutions.