
Problem 1. Find all prime numbers $a, b, c$ and positive integers $k$ which satisfy the equation

$$
a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1
$$


# Solution: 

The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies

$$
a^{2}+b^{2}+16 \cdot c^{2} \equiv 1(\bmod 3) \Leftrightarrow a^{2}+b^{2}+c^{2} \equiv 1(\bmod 3)
$$

Since $a^{2} \equiv 0,1(\bmod 3), \quad b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, 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

$$
\begin{aligned}
& a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1 \Leftrightarrow 9 \cdot k^{2}-16 \cdot c^{2}=17 \Leftrightarrow(3 k-4 c) \cdot(3 k+4 c)=17, \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 3 k - 4 c = 1 , } \\
{ 3 k + 4 c = 1 7 , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
c=2, \\
k=3,
\end{array} \text { and } \quad(a, b, c, k)=(3,3,2,3) .\right.\right.
\end{aligned}
$$

Case 2. $c=3$

If $\left(3, b_{0}, c, k\right)$ is a solution of the given equation, then $\left(b_{0}, 3, c, k\right)$ is a solution too.

Let $a=3$. We have

$$
a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1 \Leftrightarrow 9 \cdot k^{2}-b^{2}=152 \Leftrightarrow(3 k-b) \cdot(3 k+b)=152 .
$$

Both factors shall have the same parity and we obtain only 2 cases:

- $\left\{\begin{array}{l}3 k-b=2, \\ 3 k+b=76,\end{array} \Leftrightarrow\left\{\begin{array}{l}b=37, \\ k=13,\end{array}\right.\right.$ and $(a, b, c, k)=(3,37,3,13)$;
- $\left\{\begin{array}{l}3 k-b=4, \\ 3 k+b=38,\end{array} \Leftrightarrow\left\{\begin{array}{l}b=17, \\ k=7,\end{array}\right.\right.$ and $(a, b, c, k)=(3,17,3,7)$.

So, the given equation has 5 solutions:

$\{(37,3,3,13),(17,3,3,7),(3,37,3,13),(3,17,3,7),(3,3,2,3)\}$.
