5. Problem: Find all triples $(a, b, n)$ of positive integers such that $a$ and $b$ are both divisors of $n$, and $a+b=\frac{n}{2}$.
Solution: (James Xu)
Since $a$ and $b$ are both factors of $n$, we can find positive integers $x$ and $y$ such that $a=\frac{n}{x}$ and $b=\frac{n}{y}$. Then $\frac{n}{x}+\frac{n}{y}=\frac{n}{2}$ so
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{2} .
$$

Without loss of generality assume $a \leqslant b$. So $x \geqslant y$ and $\frac{1}{x} \leqslant \frac{1}{y}$ and thus
$$
\frac{2}{y} \geqslant \frac{1}{x}+\frac{1}{y}=\frac{1}{2}
$$

Therefore $y \leqslant 4$. So we try $y=1,2,3,4$ one-by-one.
- Case $1 y=1$ :
This implies $\frac{1}{x}=\frac{1}{2}-\frac{1}{y}=0$ thus $x=-2$. This doesn't work because $a$ (and therefore $x$ ) must be positive.
- Case $2 y=2$ :
This implies $\frac{1}{x}=\frac{1}{2}-\frac{1}{y}=0$ which is not possible.
- Case $3 y=3$ :
This implies $\frac{1}{x}=\frac{1}{2}-\frac{1}{y}=\frac{1}{6}$ and thus $x=6$. This means that $n$ must be a multiple of 6 . Now let $n=6 k$, and we get
$$
a=\frac{n}{x}=\frac{6 k}{6}=k \quad \text { and } \quad b=\frac{n}{y}=\frac{6 k}{3}=2 k .
$$

This yields the family of solutions $(a, b, n)=(k, 2 k, 6 k)$ where $k$ is any positive integer.
- Case $4 y=4$ :

This implies $\frac{1}{x}=\frac{1}{2}-\frac{1}{y}=\frac{1}{4}$ and thus $x=4$. This means that $n$ must be a multiple of 4 . Now let $n=4 k$, and we get
$$
a=\frac{n}{x}=\frac{4 k}{4}=k \quad \text { and } \quad b=\frac{n}{y}=\frac{4 k}{4}=k .
$$

This yields the family of solutions $(a, b, n)=(k, k, 4 k)$ where $k$ is any positive integer.
Remember that we assumed $a \leqslant b$ so we still need to swap the roles of $a$ and $b$ for our final answer. So in summary, the triples $(a, b, n)$ which satisfy the given equation are
$$
(k, 2 k, 6 k),(2 k, k, 6 k) \text { and }(k, k, 4 k)
$$
where $k$ is any positive integer.