1. Find all pairs of natural numbers $m$ and $n$ for which the quadratic equation

$$
2007 x^{2} + m n x + n = 0
$$

has exactly one solution. For each such pair, also write down the solution.
II/1. The equation has a double root if and only if the discriminant is equal to 0, that is

$$
0=(m n)^{2}-4 \cdot n \cdot 2007=n\left(m^{2} n-4 \cdot 2007\right)
$$

From this it follows that $m^{2} n=4 \cdot 2007=2^{2} \cdot 3^{2} \cdot 223$. Since 223 is a prime number, it must be a divisor of the number $n$. Let us write $n=223 n^{\prime}$. Then we have

$$
m^{2} n^{\prime}=(2 \cdot 3)^{2}
$$

We consider four possibilities. The number $m$ can be equal to $1, 2, 3$, or 6. In the first case, $n=4 \cdot 2007$, and the equation is $2007\left(x^{2}+4 x+4\right)=0$ with a double root $x=-2$. For $m=2$, we get $n=2007$ and $2007\left(x^{2}+2 x+1\right)=0$, yielding $x=-1$. For $m=3$, $n=4 \cdot 223$ and the equation is $223\left(9 x^{2}+12 x+4\right)=0$, so $x=-\frac{2}{3}$. Finally, for $m=6, n=223$ and $223\left(9 x^{2}+6 x+1\right)=0$, thus $x=-\frac{1}{3}$.
Writing that the discriminant is equal to 0 ..... 1 point
Factorization into prime factors ..... 1 point
Conclusion that 223 is a divisor of the number $n$ ..... 1 point
Considering the cases $m=1, m=2, m=3$ and $m=6$ and calculating the corresponding solutions
$-2, -1, -\frac{2}{3}$ and $-\frac{1}{3}$