## Task A-2.1.

Determine all natural numbers $n$ and prime numbers $p$ such that

$$
6 n^{2}-(2 p+9) n+p+3=0
$$
## First Solution.

Let's solve the quadratic equation for $n$:

$$
\begin{aligned}
n_{1,2} & =\frac{2 p+9 \pm \sqrt{(2 p+9)^{2}-4 \cdot 6 \cdot(p+3)}}{12} \\
& =\frac{2 p+9 \pm \sqrt{(2 p+3)^{2}}}{12} \\
& =\frac{2 p+9 \pm(2 p+3)}{12}, \\
n_{1} & =\frac{4 p+12}{12}=\frac{p}{3}+1, \quad n_{2}=\frac{6}{12}=\frac{1}{2}
\end{aligned}
$$

Since $n$ is a natural number, $n_{2}$ is not a valid solution. 1 point

For $n_{1}$, we get the condition that $\frac{p}{3}+1$ must be a natural number, and the only prime number divisible by 3 is $p=3$. 1 point The only solution to the equation is $n=2, p=3$. 1 point