Junior Balkan Mathematical Olympiad

The relation $9k^2+1\equiv 1(\mathrm{mod}3)$ implies $$a^2+b^2+16c^2\equiv 1(\mathrm{mod}3)\iff a^2+b^2+c^2\equiv 1(\mathrm{mod}3).$$ Since $a^2\equiv 0,1(\mathrm{mod}3)$, $b^2\equiv 0,1(\mathrm{mod}3)$, $c^2\equiv 0,1(\mathrm{mod}3)$, we have:

$a^2$	0	0	0	0	1	1	1	1
$b^2$	0	0	1	1	0	0	1	1
$c^2$	0	1	0	1	0	1	0	1
$a^2+b^2+c^2$	0	1	1	2	1	2	2	0

From the previous table it follows that two of three prime numbers $a$, $b$, $c$ are equal to $3$.

Case 1. $a=b=3$. We have $$a^2+b^2+16c^2=9k^2+1\iff 9k^2-16c^2=17\iff (3k-4c)(3k+4c)=17$$ If , then and $(a,b,c,k)=(3,3,2,3)$. If , then 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 $$a^2+b^2+16c^2=9k^2+1\iff 9k^2-b^2=152\iff (3k-b)(3k+b)=152.$$ Both factors shall have the same parity and we obtain only 4 cases: If , then and $(a,b,c,k)=(3,37,3,13)$. If , then and $(a,b,c,k)=(3,17,3,7)$. If , then and $(a,b,c,k)=(3,37,3,-13)$. If , then 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: $$S=\{(37,3,3,13),(17,3,3,7),(37,3,3,-13),(17,3,3,-7),(3,37,3,13),(3,17,3,7),(3,37,3,-13),(3,17,3,-7),(3,3,2,3),(3,3,2,-3)\}.$$