
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 domain of the real numbers.


Solution. Expressing $x^{n}$ from both equations

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

and comparing, we see that $2007-s x=2008-t x$, which implies that the common root can exist only for $s \neq t$ and must equal $x=1 /(t-s)$. This $x$ will be a common root of the two equations if and only if it is a root of one of them; substituting it e.g. into the first equation thus yields, upon a small manipulation, the equivalent condition

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

Since both factors on the left-hand side are integers, they must be the numbers 1 and -1 (in some order), whence in any case $t-s= \pm 1$.

a) If $t-s=1$, then the last equation reads $s-2007(t-s)=-1$. The two equations

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

with unknowns $s, t$ have 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 we find similarly as in a) the solution $s=(-1)^{n}-2007$ and $t=(-1)^{n}-2008$. (The common root is $x=-1$.)

Conclusion: The problem has exactly two solutions $(s, t)=(2006,2007)$ and $(s, t)=\left((-1)^{n}-2007,(-1)^{n}-2008\right)$.
