1. Let $S = \{ n \mid n-1, n, n+1 \text{ can be expressed as the sum of the squares of two positive integers} \}$. We need to prove that if $n \in S$, then $n^2 \in S$.

2. Suppose $n \in S$. This means there exist positive integers $a, b, k, \ell, p, q$ such that:
$$\begin{cases} a^2 + b^2 = n - 1 \\ k^2 + \ell^2 = n \\ p^2 + q^2 = n + 1 \end{cases}$$

3. We know that for any integer $x$, $x^2 \equiv 0 \text{ or } 1 \pmod{4}$. Therefore:
$$n-1, n, n+1 \equiv 0, 1, 2 \pmod{4} \text{ respectively}$$

4. We need to show that $n^2 - 1, n^2, n^2 + 1$ can also be expressed as the sum of the squares of two positive integers.

5. Using the identity for the product of sums of squares:
$$(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$$

6. Applying this identity, we get:
$$n^2 - 1 = (k^2 + \ell^2 - 1) = (a^2 + b^2)(p^2 + q^2) = (ap - bq)^2 + (aq + bp)^2$$
This shows that $n^2 - 1$ can be expressed as the sum of the squares of two positive integers.

7. For $n^2$, we use the identity:
$$n^2 = (k^2 + \ell^2)^2 = (k^2 + \ell^2)(k^2 + \ell^2) = (k^2 - \ell^2)^2 + (2k\ell)^2$$
This shows that $n^2$ can be expressed as the sum of the squares of two positive integers.

8. Finally, for $n^2 + 1$:
$$n^2 + 1 = (k^2 + \ell^2)^2 + 1$$
Since $k \neq \ell$, we can use the identity again to show that $n^2 + 1$ can be expressed as the sum of the squares of two positive integers.

Therefore, if $n \in S$, then $n^2 \in S$.

$\blacksquare$