5. Given a natural number $n$ greater than 1, find all pairs of integers $a$ and $b$ such that the two equations $x^{n}+a x-2008=0$ and $x^{n}+b x-2009=0$ have at least one common real root.
Solution:

Subtracting both equations, we have $(b-a) x=1$. Then, if these equations are to have a common root, it must be $x=1/(b-a)$. Note that $a$ cannot be equal to $b$. Substituting into one of the equations, we will have

$$
(b-a)^{n-1}(a-2008(b-a))=-1
$$

and since $a$ and $b$ are integers, these two factors will be one equal to $+1$ and the other equal to -1.

If $(b-a)=1$, we get $a=-1+2008=2007$, and thus $b=2008$.

If $(b-a)=-1$, we get $a=(-1)^{n-1}-2008$, and thus $b=(-1)^{n-1}-2009$.

Therefore, the only pairs of numbers $(a, b)$ are

$$
(2007,2008) \text { and }\left((-1)^{n-1}-2008,(-1)^{n-1}-2009\right)
$$