14.3.4 * Find all positive integers $n>1$, such that any of its divisors greater than 1 has the form $a^{r}+1$, where $a$ is a positive integer, and $r$ is a positive integer greater than 1.
If $p$ is a prime factor of $n$, then $p=a^{r}+1$. When $r$ is odd, $a^{r}+1$ has a factor $a+1$, which contradicts $p$ being a prime, unless $p=2, a=1$. Therefore, we can always assume $r$ is even, and $r=2$ (otherwise, replace $a$ with $a^{\frac{r}{2}}$), i.e., $p=a^{2}+1$. If $n$ has a factor $pq$, where $p, q$ are primes (which can be the same), then $p=a^{2}+1, q=b^{2}+1$, and $pq=c^{r}+1$. When $r$ is odd, $c^{r}+1$ has a factor $c+1$, so $p=c+1, c=a^{2}$. Thus, we can still set $r=2$ (otherwise, replace $c$ with $c^{\frac{1}{2}}$). Assuming without loss of generality that $a \leqslant b$, from $\left(a^{2}+1\right)\left(b^{2}+1\right)=c^{2}+1$, we get $a^{2}\left(b^{2}+1\right)=(c-b)(c+b)$. (1) Since $b^{2}+1$ is a prime, $b^{2}+1$ divides $c-b$ or $c+b$. If $b^{2}+1 \leqslant c-b$, then $c+b > b^{2}+1 > a^{2}$, (1) cannot hold. If $b^{2}+1 \leqslant \frac{1}{2}(c+b)$, then $c-b \geqslant 2\left(b^{2}+1\right)-2 b=b^{2}+1+(b-1)^{2}>a^{2}$, (1) also cannot hold. Therefore, we must have $\left\{\begin{array}{l}b^{2}+1=c+b, \\ a^{2}=c-b .\end{array}\right.$ Solving gives $c=b^{2}-b+1, a=b-1$. Thus, $b^{2}+1$, $(b-1)^{2}+1$ are both primes. But $b, b-1$ have different parities, and only when $b=2$ is it possible for $b^{2}+1$ and $(b-1)^{2}+1$ to both be primes. Therefore, $n=2 \times 5=10$ or $n$ itself is a prime of the form $a^{2}+1$ (which clearly satisfies the requirement).

Comment: Are there infinitely many primes of the form $a^{2}+1$? This is still an unsolved problem. Currently, it is only known that there are infinitely many numbers of the form $a^{2}+1$ that are either primes or the product of two primes.