Solution:

We shall use the following lemma.

Lemma. Let $x$, $y$ and $n$ be positive integers such that $\frac{xy}{x+y} > n$. Then
$$\frac{xy}{x+y} \geq n + \frac{1}{n^{2} + 2n + 2}$$
with equality if and only if $\{x, y\} = \left\{n+1, n^{2}+n+1\right\}$.

Proof of the lemma. Since $xy > n(x+y)$, we have $xy = n(x+y) + r$, where $r$ is a positive integer. Then $(x-n)(y-n) = n^{2} + r$, which implies that $x > n$ and $y > n$. Set $x = n + d_{1}$ and $y = n + d_{2}$. Then $d_{1} d_{2} = n^{2} + r$. Now using the inequalities $\frac{r}{A+r} \geq \frac{1}{A+1}$ and $d_{1} + d_{2} \leq 1 + n^{2} + r$ (the latter follows from $d_{1} + d_{2} \leq 1 + d_{1} d_{2}$), we get
$$\begin{aligned} \frac{xy}{x+y} &= \frac{n^{2} + d_{1} d_{2} + n(d_{1} + d_{2})}{2n + d_{1} + d_{2}} = n + \frac{r}{2n + d_{1} + d_{2}} \\ &\geq n + \frac{r}{2n + n^{2} + 1 + r} \geq n + \frac{1}{n^{2} + 2n + 2} \end{aligned}$$
Note that the equality is attained if and only if $\{x, y\} = \left\{n+1, n^{2}+n+1\right\}$. This completes the proof of the lemma.

The condition of the problem implies that $c\left(c^{2}-c+1\right) = p a b$ and $a+b = q\left(c^{2}+1\right)$, where $p$ and $q$ are positive integers. Therefore
$$\frac{c\left(c^{2}-c+1\right)}{c^{2}+1} = \frac{pqab}{a+b} = \frac{xy}{x+y}$$
where $x = pqa$ and $y = pqb$. Then
$$\frac{xy}{x+y} = c - \frac{c^{2}}{c^{2}+1} = c-1 + \frac{1}{c^{2}+1}$$
whence $\frac{xy}{x+y} > c-1$. Now the lemma gives
$$\frac{xy}{x+y} \geq c-1 + \frac{1}{(c-1)^{2} + 2(c-1) + 2} = c-1 + \frac{1}{c^{2}+1}$$
Hence we have the case of equality and therefore
$$\{x, y\} = \left\{c, c^{2}-c+1\right\}$$
Since the numbers $c$ and $c^{2}-c+1$ are coprime and $x = pqa$, $y = pqb$, it follows that $p = q = 1$. Hence $\{a, b\} = \left\{c, c^{2}-c+1\right\}$.