A sequence satisfying the given condition exists. The minimal common difference is 84, and one such sequence is
29, 113, 197, 281, 365, 449, 533, 617, 701, 785.
We need the following lemmas.
Lemma 1. An integer of the form $n^2 + 1$ has no prime factor of the form $4k + 3$.
Proof. This follows immediately by using Fermat's little theorem.

Lemma 2. If $a_1, b_1, a_2, b_2$ are nonzero integers such that $a_i$ and $b_i$ are coprime ($i = 1, 2$), then $(a_1^2 + b_1^2)(a_2^2 + b_2^2)$ can also be expressed as a sum of squares of two coprime integers.
Proof. Note that $(a_1^2 + b_1^2)(a_2^2 + b_2^2) = (a_1a_2 - b_1b_2)^2 + (a_1b_2 + a_2b_1)^2$. We prove by contradiction that
$$\gcd(a_1a_2 - b_1b_2, a_1b_2 + a_2b_1) = 1.$$
Suppose to the contrary that there exists a prime $p$ dividing both $a_1a_2 - b_1b_2$ and $a_1b_2 + a_2b_1$. Then we have
$$p \mid a_1(a_2^2 + b_2^2) \quad \text{and} \quad p \mid b_1(a_2^2 + b_2^2).$$
Since $a_1, b_1$ are coprime, we obtain that one of the following (i), (ii) must hold:
(i) $p \mid a_1$ and $p \mid (a_2^2 + b_2^2)$
(ii) $p \mid b_1$ and $p \mid (a_2^2 + b_2^2)$
Further noting that $a_2, b_2$ are coprime, we obtain that one of the following (i)*, (ii)* must hold:
(i)* $p \mid a_1$ and $p$ does not divide $b_1, a_2, b_2$
(ii)* $p \mid b_1$ and $p$ does not divide $a_1, a_2, b_2$
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From (i)*, we get that $p$ does not divide $a_1a_2-b_1b_2$; from (ii)*, we get that $p$ does not divide $a_1b_2+a_2b_1$. Either case leads to a contradiction. Hence we have proved that $\text{gcd}(a_1a_2 - b_1b_2, a_1b_2 + a_2b_1) = 1$, and this completes the proof of the lemma.

Lemma 3. If $a, b$ are coprime integers, then there must exist an integer $m$ such that $a^2 + b^2 \mid m^2 + 1$.
Proof. Since $a, b$ are coprime, there must exist integers $c, d$ such that $ac - bd = 1$. Then
$$a^2 + b^2 = A \mid (a^2 + b^2)(c^2 + d^2) = (ad + bc)^2 + (ac - bd)^2 = m^2 + 1,$$
where $m = ad + bc$. This completes the proof.

Now we return to the original problem. We discuss this in several steps:

Claim 1. $a_i \equiv 1 \pmod 4$, $i = 1, 2, \dots, 10$.
Proof. We discuss two cases.
(i) If $a_1$ is even, then $a_3, a_5, a_7, a_9$ are also even. From this we obtain $32 \mid n^2 + 1$, which is clearly impossible.
(ii) If $a_1$ is odd, then since $a_1 \mid n^2 + 1$, by Lemma 1 we know that $a_1 \equiv 1 \pmod 4$ must hold.
The same argument gives $a_2 \equiv 1 \pmod 4$. Because $a_1, a_2, \dots, a_{10}$ form an arithmetic sequence, the common difference must be a multiple of 4, hence Claim 1 is proved.

Claim 2. Let the common difference be $d$. Then $84 \mid d$.
Proof. From Claim 1 we obtain $4 \mid d$. We now prove that $21 \mid d$.
If $d$ is not a multiple of 3, then $\{a_1, a_2, a_3\}$ forms a complete residue system modulo 3, from which we obtain $3 \mid n^2 + 1$, which is clearly impossible. Similarly, if $d$ is not a multiple of 7, then $\{a_1, a_2, \dots, a_7\}$ forms a complete residue system modulo 7, from which we can also obtain $7 \mid n^2 + 1$, which is likewise impossible. Hence $21 \mid d$ is proved, and Claim 2 is proved.

Now we construct an arithmetic sequence with common difference 84 satisfying the condition of the problem. Moreover, by Lemma 1 we obtain that every prime factor of each $a_i$ must be congruent to 1 modulo 4. We have the following result:

Claim 3. $a_1 \equiv 3 \pmod{4}$ or $-4 \pmod{11}$. This only requires knowing from Lemma 1 that 11 cannot divide $n^2+1$.
From Claim 1 and Claim 3, and noting that $a_1$ is not divisible by 3 or 7, we sieve out the values of $a_1$ in increasing order as 25, 29, 47, 73, ...
When $a_1 = 25$, $a_5 = 25 + 84 \times 4 = 361 = 19^2$. Since $19 \equiv 3 \pmod 4$, this does not satisfy the requirement (it contradicts Lemma 1).
When $a_1 = 29 = 2^2 + 5^2$, we can obtain:
$$-a_2 = 29 + 84 = 113 = 7^2 + 8^2 \text{ and is prime;}$$
$$-a_3 = 29 + 84 \times 2 = 197 = 1^2 + 14^2 \text{ and is prime;}$$
$$-a_4 = 29 + 84 \times 3 = 281 = 5^2 + 16^2 \text{ and is prime;}$$
$$-a_5 = 29 + 84 \times 4 = 365 = (1^2 + 2^2)(3^2 + 8^2) = 2^2 + 19^2$$
$$-a_6 = 29 + 84 \times 5 = 449 = 7^2 + 20^2 \text{ and is prime;}$$
$$-a_7 = 29 + 84 \times 6 = 533 = (2^2 + 3^2)(4^2 + 5^2) = 7^2 + 22^2;$$
$$-a_8 = 29 + 84 \times 7 = 617 = 16^2 + 19^2 \text{ and is prime;}$$
$$-a_9 = 29 + 84 \times 8 = 701 = 5^2 + 26^2 \text{ and is prime;}$$
$$-a_{10} = 29 + 84 \times 9 = 785 = (1^2 + 2^2)(6^2 + 11^2) = 1^2 + 28^2.$$

$$\prod_{i=1}^{10} a_i \mid n^2 + 1.$$
From the above argument we also obtain that \{29, 113, 197, 281, 365, 449, 533, 617, 701, 785\} is
a sequence satisfying the condition of the problem with "minimal" common difference.