$29 \cdot 34$ To make the polynomial $x^{2}-2 x-n$ factorable into the product of two linear polynomials with integer coefficients, the number of natural numbers $n$ not exceeding 100 is
(A) 8.
(B) 9.
(C) 10.
(D) 11.
(1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989.)
[Solution] Let $x^{2}-2 x-n=(x-a)(x-b)$, then $a, b$ have different signs (product is $-n$), without loss of generality, assume $a>0, b<0$.
From $a+b=2, a b=-n$, then $n=a(a-2)=(a-1)^{2}-1$.
Let $a-1=c$, then $n$ can be written in the form of $c^{2}-1$, where $c$ is a non-negative integer.
Among the natural numbers not greater than 100, the perfect squares are $1,4,9, \cdots, 81,100$, a total of 10, so $n$ can take the natural numbers $3,8, \cdots, 80,99$, a total of 9.
Therefore, the answer is $(B)$.