1. Let $n$ be a given natural number greater than 1. Find all pairs of integers $s$ and $t$ for which the equations

$$
x^{n}+s x-2007=0, \quad x^{n}+t x-2008=0
$$

have at least one common root in the set of real numbers.
1. By expressing the term $x^{n} \mathrm{z}$ in both equations

$$
x^{n}=2007-s x, \quad x^{n}=2008-t x
$$

we obtain the equation $2007-s x=2008-t x$ by comparison, according to which a common root $x$ can only exist in the case $s \neq t$ and must be of the form $x=1 /(t-s)$. Such an $x$ is indeed a root of both original equations if and only if it is a root of one of them; by substituting, for example, into the first equation, we get the equivalent condition after rearrangement

$$
(t-s)^{n-1} \cdot(s-2007(t-s))=-1
$$

Since both factors on the left side are integers, they must be the numbers 1 and -1 in some order, so according to the first factor, we must also have $t-s= \pm 1$.

a) If $t-s=1$, then the found condition has the form $s-2007(t-s)=-1$. The system of equations for the unknown values $s, t$

$$
t-s=1, \quad s-2007(t-s)=-1
$$

has the unique solution $s=2006$ and $t=2007$. (The common root is $x=1$.)

b) If $t-s=-1$, then $s-2007(t-s)=(-1)^{n}$, from which, similarly to case a), we find the solution $s=(-1)^{n}-2007$ and $t=(-1)^{n}-2008$. (The common root is $x=-1$.)

Conclusion: The conditions of the problem are satisfied precisely by the two pairs $(s, t)=(2006,2007)$ and $(s, t)=\left((-1)^{n}-2007,(-1)^{n}-2008\right)$.

For a complete solution, award 6 points. Award 2 points for deriving the formula $x=1 /(t-s)$, and another 2 points for subsequently determining $t-s= \pm 1$. Finally, award 1 point each for determining the pairs $(s, t)$ in the individual cases a) and b). The common root $x$ is a real number; if the solver does not justify that it is an integer, then for an otherwise complete solution based on the argument that the number $x$ as a common divisor of the numbers 2007 and 2008 must be equal to $\pm 1$, award only 1 point, and similarly if the solver somehow guesses both pairs $(s, t)$ (award no points for just one of them). Check whether the protocols aspiring to completeness lack a verification (which is not necessary in our explanation due to the mentioned equivalence), and penalize the absence of verification by a loss of 1 point.