Solution:

From the first equation we obtain $c = a + b - n$; substituting into the second equation we get
$$n = a^{2} + b^{2} - (a + b - n)^{2} = -2 a b + 2 a n + 2 b n - n^{2}.$$
Rearranging, we obtain
$$(a - n)(b - n) = a b - n a - n b + n^{2} = \frac{n^{2} - n}{2} = \frac{n(n-1)}{2}$$
The right-hand side is always an integer, because one of $n$ and $n-1$ is always even. The solutions of the equation correspond to the ways of factoring $\frac{n(n-1)}{2} = x y$ as a product of two integers $x$ and $y$. Since $n \neq 0,1$ by hypothesis, $\frac{n(n-1)}{2}$ is a positive integer, and hence the number of these factorizations is finite; it follows that the system will have finitely many solutions, $(a, b, c) = (x + n, y + n, x + y + n)$.
To find at least one solution, for every $n$ we can choose $x = 1, y = \frac{n(n-1)}{2}$, from which we obtain the solution
$$(a, b, c) = \left(n + 1, \frac{n(n+1)}{2}, \frac{n(n+1)}{2} + 1\right).$$