Three, let $n$ be an integer of the form $a^{2}+b^{2}$, where integers $a, b$ are coprime. If $p$ is a prime, $p<\sqrt{n}$, then $p \mid a b$. Determine all such $n$.

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Three, first note that $n$ is a prime number, so $n=2$ or $n$ is a number of the form $4k+1$. Below we prove that $n \leq 5^2 = 25$.
Assume $n > 25$, then all primes not greater than 5 divide $ab$, and
$$
\begin{aligned}
n & =a^2 + b^2 \geq c^2 + d^2 \geq 2cd \\
& \geq 2(2 \times 3 \times 5) = 60 > 49 = 7^2.
\end{aligned}
$$

Here, $c$ and $d$ are the products of the distinct prime factors of $a$ and $b$, respectively.
Let $p_k$ be the $k$-th prime number. We will prove by mathematical induction that:
If $p_k \geq 5$, then $2(2 \times 3 \times \cdots \times p_k) \geq p_{k+1}^2$.
We know that $2p_{k+1} > p_{k+2}$ (Bertrand's postulate). By the induction hypothesis, we have
$$
\begin{array}{l}
2(2 \times 3 \times \cdots \times p_k \times p_{k+1}) > p_{k+1}^3 \\
> p_{k+1} \cdot \left(\frac{p_{k+2}}{2}\right)^2 > p_{k+2}^2,
\end{array}
$$

where $p_{k+1} > 5$. Therefore, the proposition is correct. From this proposition and the assumption $n > 25$, we can conclude: for any prime $p$, $n > p^2$, which is clearly not possible. Hence, $n \leq 25$.
The possible values of $n$ can only be $2, 5, 13$, and 17. However, $17 = 4^2 + 1$, and $3 \leq \sqrt{17}$, but 3 does not divide 2 or 1, while $2 = 1^2 + 1^2$, $5 = 2^2 + 1$, $13 = 2^2 + 3^2$, and they satisfy the assumption conditions, so $n = 2, 5$, or 13.