Junior Balkan Mathematical Olympiad

The relation $9k^2 + 1 \equiv 1 \pmod{3}$ implies
$$a^2 + b^2 + 16c^2 \equiv 1 \pmod{3} \quad \Leftrightarrow \quad a^2 + b^2 + c^2 \equiv 1 \pmod{3}.$$
Since $a^2 \equiv 0,1 \pmod{3}$, $b^2 \equiv 0,1 \pmod{3}$, $c^2 \equiv 0,1 \pmod{3}$, we have:

Case 1. $a = b = 3$. We have
$$a^2 + b^2 + 16c^2 = 9k^2 + 1 \Leftrightarrow 9k^2 - 16c^2 = 17 \Leftrightarrow (3k - 4c)(3k + 4c) = 17$$
If $\begin{cases} 3k - 4c = 1 \\ 3k + 4c = 17 \end{cases}$, then $\begin{cases} c = 2 \\ k = 3 \end{cases}$ and $(a, b, c, k) = (3, 3, 2, 3)$.
If $\begin{cases} 3k - 4c = -1 \\ 3k + 4c = -17 \end{cases}$, then $\begin{cases} c = 2 \\ k = -3 \end{cases}$ and $(a, b, c, k) = (3, 3, 2, -3)$.

Case 2. $c = 3$. If $(3, b_0, c, k)$ is a solution of the given equation, then $(b_0, 3, c, k)$ is a solution, too.
Let $a = 3$. We have
$$a^2 + b^2 + 16c^2 = 9k^2 + 1 \Leftrightarrow 9k^2 - b^2 = 152 \Leftrightarrow (3k - b)(3k + b) = 152.$$
Both factors shall have the same parity and we obtain only 4 cases:
If $\begin{cases} 3k - b = 2 \\ 3k + b = 76 \end{cases}$, then $\begin{cases} b = 37 \\ k = 13 \end{cases}$ and $(a, b, c, k) = (3, 37, 3, 13)$.
If $\begin{cases} 3k - b = 4 \\ 3k + b = 38 \end{cases}$, then $\begin{cases} b = 17 \\ k = 7 \end{cases}$ and $(a, b, c, k) = (3, 17, 3, 7)$.
If $\begin{cases} 3k - b = -76 \\ 3k + b = -2 \end{cases}$, then $\begin{cases} b = 37 \\ k = -13 \end{cases}$ and $(a, b, c, k) = (3, 37, 3, -13)$.
If $\begin{cases} 3k - b = -38 \\ 3k + b = -4 \end{cases}$, then $\begin{cases} b = 17 \\ k = -7 \end{cases}$ and $(a, b, c, k) = (3, 17, 3, -7)$.
In addition, $(a, b, c, k) \in \{(37, 3, 3, 13), (17, 3, 3, 7), (37, 3, 3, -13), (17, 3, 3, -7)\}$.
So, the given equation has 10 solutions:
$$S = \{(37, 3, 3, 13), (17, 3, 3, 7), (37, 3, 3, -13), (17, 3, 3, -7), (3, 37, 3, 13), (3, 17, 3, 7), (3, 37, 3, -13), (3, 17, 3, -7), (3, 3, 2, 3), (3, 3, 2, -3)\}.$$