Solution:

a.
First, we observe that $d_{4}$ cannot be less than or equal to $\sqrt{n}$, because otherwise $d_{3}^{2} + d_{4}^{2}$ would be less than $(\sqrt{n} - 1)^{2} + \sqrt{n}^{2} < 2n + 1$, while we know it equals $2n + 1$. Likewise, if $d_{3}$ were greater than or equal to $\sqrt{n}$ we would have $d_{3}^{2} + d_{4}^{2} > 2n + 1$, again a contradiction: hence $d_{3} < \sqrt{n} < d_{4}$, and in particular $\sqrt{n}$ is not a divisor of $n$.

We further observe that if $d$ is a divisor of $n$ then so is $n / d$, and this operation exchanges the divisors greater than $\sqrt{n}$ with those less than $\sqrt{n}$. It follows that $n$ has as many divisors greater than $\sqrt{n}$ as divisors less than $\sqrt{n}$: since these are 3 by what we have just said, $n$ has 6 divisors.

b.
First solution
Note that we also have $d_{3} d_{4} = n$: the smallest divisor greater than $\sqrt{n}$ (that is, $d_{4}$) equals $n / d$, where $d$ is the largest divisor less than $\sqrt{n}$ (that is, $d_{3}$). From the equation $d_{3}^{2} + d_{4}^{2} = 2n + 1$ we then deduce $d_{3}^{2} + d_{4}^{2} = 2 d_{3} d_{4} + 1$, that is, $(d_{3} - d_{4})^{2} = 1$, and hence $d_{4} - d_{3} = 1$ (since $d_{4} > d_{3}$). The number $n$ can then be written as $n = d_{3}(d_{3} + 1)$, and in particular it is even, because one of the two factors $d_{3}, d_{3} + 1$ is even. The divisors of $n$ are therefore $1, 2, d_{3}, d_{3} + 1, \frac{n}{2}, n$.

Since the divisors of $d_{3}$ are also divisors of $n$, either $d_{3}$ has no divisors other than 1 and $d_{3}$ (and hence is prime), or its only nontrivial divisor is 2 (and hence $d_{3} = 4$). In the second case we have $n = d_{3}(d_{3} + 1) = 4 \cdot 5 = 20$, and otherwise we repeat the same reasoning with the divisors of $d_{3} + 1$: if $d_{3}$ is prime then $d_{3} + 1$ is even, and on the other hand it cannot have nontrivial divisors other than 2, so $d_{3} + 1 = 4$ and $n = d_{3}(d_{3} + 1) = 12$.

Finally, one easily checks that $n = 12$ and $n = 20$ are indeed solutions (in the two cases we have $d_{3}^{2} + d_{4}^{2} = 3^{2} + 4^{2} = 25 = 2 \cdot 12 + 1$, $d_{3}^{2} + d_{4}^{2} = 16 + 25 = 41 = 2 \cdot 20 + 1$), and by what we have already said these are the only ones.

Second solution
Exactly one of $d_{3}$ and $d_{4}$ is even: if $d_{3}, d_{4}$ had the same parity, then $d_{3}^{2} + d_{4}^{2} = 2n + 1$ would be even, which it clearly is not. Moreover $d_{3} d_{4} = n$: the smallest divisor greater than $\sqrt{n}$ (that is, $d_{4}$) equals $n / d$, where $d$ is the largest divisor less than $\sqrt{n}$ (that is, $d_{3}$). The divisors of $n$ are therefore $1, 2, d_{3}, n / d_{3}, n / 2, n$.

Since the divisors of $d_{3}$ are also divisors of $n$, either $d_{3}$ has no divisors other than 1 and $d_{3}$ (and hence is prime), or its only nontrivial divisor is 2 (and hence $d_{3} = 4$). Similarly, the only possible divisors of $d_{4}$ are $1, 2, d_{3}$, but if $d_{3}$ divided $d_{4}$ then $d_{3}$ would divide $d_{3}^{2}, d_{4}^{2}$ and $n$ and hence, by subtraction, would divide $d_{3}^{2} + d_{4}^{2} - 2n = 1$, absurd. So $d_{4}$ too is either prime or equal to 4. If $d_{4} = 4$, then we have $2 = d_{2} < d_{3} < d_{4} = 4$ and hence $d_{3} = 3$, $n = 12$. Otherwise $d_{3} = 4$, $d_{4}$ is a prime $p$, and $n = 4p$. The equation of the problem then becomes $4^{2} + p^{2} = 8p + 1$, whose solutions are $p = 5, p = 3$. Since $p = d_{4} > d_{3} = 4$, the only possibility is $p = 5$, which gives the other solution $n = 20$.