Consider an arbitrary integer $k$. Take a complex number $\alpha$ that meets $\alpha^2 = k$. (For example, let $\alpha = \sqrt{k}$ if $k$ is nonnegative, and $\alpha = i\sqrt{-k}$ if $k$ negative.)
It is easy verify the following equalities:
$$(n + \alpha)(n + 1 - \alpha) = ((n(n + 1) - k) + \alpha), \\ (n - \alpha)(n + 1 + \alpha) = ((n(n + 1) - k) - \alpha).$$
By multiplying these two formulae, we obtain an identity,
$$(n^2 - k)((n + 1)^2 - k) = ((n(n + 1) - k)^2 - k).$$
In consequence, for any $k$, we can pick up any $n$ and let $(a, b, c) = (n, n + 1, n(n + 1) - k)$ so that the required equality holds. There are infinitely many $n$ and hence $(a, b, c)$. Therefore, every integer $k$ meets the required condition.