8. If $n$ is a natural number less than 50, find all values of $n$ such that the values of the algebraic expressions $4n+5$ and $7n+6$ have a common divisor greater than 1.
8. Solution: Let $(4 n+5,7 n+6)=d>1$, then $d|(4 n+5), d|(7 n+6)$.
Thus $d \mid(7 n+6-(4 n+5))=3 n+1$,
$$
\begin{array}{l}
d \mid((4 n+5)-(3 n+1))=n+4, \\
d \mid((3 n+1)-2(n+4))=n-7, \\
d \mid((n+4)-(n-7))=11 .
\end{array}
$$

Since 11 is a prime number, then $d=11$.
Let $n-7=11 k$, then
$0<n=11 k+7<50$. Solving for $k$ gives $k=0,1,2,3$.
Thus $n=7,18,29,40$.