1. Suppose there exist positive integers $x, y, z, w$ (with $x < y < z < w$) such that $x^2, y^2, z^2, w^2$ are four consecutive terms of an arithmetic sequence. We consider such $x, y, z, w$ such that $x^2 + y^2 + z^2 + w^2$ is minimized.
2. Since $x^2, y^2, z^2, w^2$ are in arithmetic progression, there exists a common difference $d$ such that:
$$y^2 = x^2 + d, \quad z^2 = y^2 + d, \quad w^2 = z^2 + d$$
This implies:
$$z^2 = x^2 + 2d \quad \text{and} \quad w^2 = x^2 + 3d$$
3. From $y^2 = x^2 + d$ and $z^2 = y^2 + d$, we get:
$$2y^2 = x^2 + z^2$$
Similarly, from $z^2 = y^2 + d$ and $w^2 = z^2 + d$, we get:
$$2z^2 = y^2 + w^2$$
4. Since $x, y, z, w$ are positive integers, $x, y, z, w$ must all be odd. This is because if any of them were even, the equations $2y^2 = x^2 + z^2$ and $2z^2 = y^2 + w^2$ would not hold for integers.
5. We can express $y$ and $z$ in terms of $x$ and $w$:
$$y = \frac{x + z}{2} \quad \text{and} \quad z = \frac{y + w}{2}$$
6. Therefore, there exist integers $m, n$ such that:
$$y = m^2 + n^2, \quad \frac{z + x}{2} = m^2 - n^2, \quad \frac{z - x}{2} = 2mn$$
and integers $p, q$ such that:
$$z = p^2 + q^2, \quad \frac{w + y}{2} = p^2 - q^2, \quad \frac{w - y}{2} = 2pq$$
7. From the minimality of $x^2 + y^2 + z^2 + w^2$, we can see that $y, \frac{z + x}{2},$ etc., are coprime. Therefore:
$$y = m^2 + n^2 = |p^2 - q^2 - 2pq|, \quad z = m^2 - n^2 + 2mn = p^2 + q^2$$
8. Case 1: $y = m^2 + n^2 = p^2 - q^2 - 2pq$
$$\frac{z + y}{2} = m^2 + mn = p^2 - pq \quad \Rightarrow \quad p^2 - m^2 = mn + pq$$
$$\frac{z - y}{2} = mn - n^2 = q^2 + pq \quad \Rightarrow \quad q^2 + n^2 = mn - pq$$
This leads to:
$$(p^2 - m^2)(q^2 + n^2) = m^2 n^2 - p^2 q^2$$
which simplifies to:
$$p^2 (n^2 + 2q^2) = m^2 (2n^2 + q^2)$$
Assuming $\gcd(m, p) = \gcd(n, q) = 1$, we have:
$$\gcd(n^2 + 2q^2, 2n^2 + q^2) = 1 \text{ or } 3$$
If $\gcd(n^2 + 2q^2, 2n^2 + q^2) = 1$, then:
$$p^2 = 2n^2 + q^2, \quad m^2 = n^2 + 2q^2$$
This implies $3 \mid p^2 + m^2$, so $3 \mid p, m$, which contradicts $\gcd(m, p) = 1$.

If $\gcd(n^2 + 2q^2, 2n^2 + q^2) = 3$, then:
$$3p^2 = 2n^2 + q^2, \quad 3m^2 = n^2 + 2q^2$$
This implies $n^2, p^2, m^2, q^2$ are four consecutive terms of an arithmetic sequence, which contradicts the minimality of $x^2 + y^2 + z^2 + w^2$.

9. Case 2: $y = m^2 + n^2 = 2pq - p^2 + q^2$
$$\frac{z + y}{2} = m^2 + mn = pq + q^2$$
$$\frac{z - y}{2} = mn - n^2 = p^2 - pq$$
This leads to:
$$(q^2 - m^2)(p^2 + n^2) = m^2 n^2 - p^2 q^2$$
which simplifies to:
$$m^2 (p^2 + 2n^2) = q^2 (2p^2 + n^2)$$
From then on, we proceed in the same way as in Case 1, finishing the proof.

$\blacksquare$