Solution:

By definition of $D_{n}$, taking $a = D_{n}$ we see that $D_{n}$ divides $D_{n}^{n} + (D_{n} + 1)^{n} + (D_{n} + 2)^{n}$, and taking $a = D_{n} + 1$ we also get that $D_{n}$ divides $(D_{n} + 1)^{n} + (D_{n} + 2)^{n} + (D_{n} + 3)^{n}$. By subtraction $D_{n}$ divides $((D_{n} + 1)^{n} + (D_{n} + 2)^{n} + (D_{n} + 3)^{n}) - (D_{n}^{n} + (D_{n} + 1)^{n} + (D_{n} + 2)^{n}) = (D_{n} + 3)^{n} - D_{n}^{n}$, and since clearly $D_{n}$ divides $D_{n}^{n}$, again by subtraction we deduce that $D_{n}$ divides $(D_{n} + 3)^{n}$. Let us now expand $(D_{n} + 3)^{n}$: it is a sum of terms of the form $D_{n}^{b} 3^{n-b}$, where $b$ ranges between $0$ and $n$, and therefore each of them is divisible by $D_{n}$, with the sole exception of the single term with $b = 0$. We can therefore write $(D_{n} + 3)^{n} = 3^{n} +$ multiples of $D_{n}$, from which we deduce that $D_{n}$ divides $3^{n}$: it follows that $D_{n}$ itself is a power of $3$, as desired.

Let us first observe that $D_{2} = 1$: indeed $D_{2}$ is a power of $3$ (by part a) and by definition divides $1^{2} + 2^{2} + 3^{2} = 14$, which forces $D_{2} = 1$.

We then have $D_{1} = 3$: taking $a = 1$ we obtain that $D_{1}$ divides $6$, hence (since it is a power of $3$) $D_{1}$ is $1$ or $3$. But on the other hand for every $a$ the number $a + (a+1) + (a+2) = 3(a+1)$ is a multiple of $3$, so $D_{1} = 3$.

We now want to show that for every $k \geq 0$ we have $D_{3^{k}} = 3^{k+1}$. For $k = 0$ we have just verified it. Since $D_{3^{k}}$ is a power of $3$, it suffices to see that $D_{3^{k}}$ is divisible by $3^{k+1}$ but not by $3^{k+2}$. Since one passes from $a^{n} + (a+1)^{n} + (a+2)^{n}$ to $(a+1)^{n} + (a+2)^{n} + (a+3)^{n}$ by adding $(a+3)^{n} - a^{n}$, to prove that $3^{k+1}$ divides $D_{3^{k}}$ it suffices to check that $3^{k+1}$ divides $1^{3^{k}} + 2^{3^{k}} + 3^{3^{k}}$ and each of the differences $(a+3)^{3^{k}} - a^{3^{k}}$. To prove that $D_{3^{k}}$ is not divisible by $3^{k+2}$ it suffices to prove that $1^{3^{k}} + 2^{3^{k}} + 3^{3^{k}}$ is not divisible by $3^{k+2}$.

Let us first deal with the statement about the differences, by induction on $k$: for $k = 0$ it is obvious. Suppose then that, for some $k$, $(a+3)^{3^{k}} - a^{3^{k}}$ is a multiple of $3^{k+1}$. We can write $(a+3)^{3^{k}} = a^{3^{k}} + c \cdot 3^{k+1}$, and raising to the cube we find $(a+3)^{3^{k+1}} = a^{3^{k+1}} + 3 a^{2 \cdot 3^{k}} \cdot c \cdot 3^{k+1} + 3 a^{3^{k}} \cdot c^{2} \cdot 3^{2(k+1)} + c^{3} \cdot 3^{3(k+1)}$, that is $(a+3)^{3^{k+1}} - a^{3^{k+1}} = a^{2 \cdot 3^{k}} \cdot c \cdot 3^{k+2} + a^{3^{k}} \cdot c^{2} \cdot 3^{2k+3} + c^{3} \cdot 3^{3k+3}$, which is clearly divisible by $3^{k+2}$, as desired.

As for $1^{3^{k}} + 2^{3^{k}} + 3^{3^{k}}$, we prove at the same time that it is divisible by $3^{k+1}$ but not by $3^{k+2}$. The statement is trivial for $k = 0$, since $1 + 2 + 3$ is divisible by $3$ but not by $9$. For $k \geq 1$ we observe that $3^{3^{k}}$ is divisible by $3^{k+2}$, so it suffices to prove the statement for $1^{3^{k}} + 2^{3^{k}}$ instead of for $1^{3^{k}} + 2^{3^{k}} + 3^{3^{k}}$. Again, we use induction on $k$. For $k = 1$ we have that $1 + 2^{3}$ is divisible by $3^{2}$ but not by $3^{3}$. Suppose now that $1 + 2^{3^{k}}$ is divisible by $3^{k+1}$ but not by $3^{k+2}$: this means that $1 + 2^{3^{k}} = 3^{k+1} \cdot c$, where $c$ is not a multiple of $3$. We then have $1 + 2^{3^{k+1}} = 1 + (2^{3^{k}})^{3} = 1 + (3^{k+1} c - 1)^{3} = 3^{k+2}(c - 3^{k+1} c^{2} + 3^{2k+2} c^{3})$. The right-hand side is a multiple of $3^{k+2}$ but not of $3^{k+3}$: indeed $(c - 3^{k+1} c^{2} + 3^{2k+2} c^{3})$ is not divisible by $3$, since it is the sum of multiples of $3$ and of $c$, which by hypothesis is not divisible by $3$. This concludes the proof.