9. How many positive integers $n$ are there such that $7 n+1$ is a perfect square and $3 n+1<2008$ ?
(A) 6
(B) 9
(C) 12
(D) 15
(E) 18
9. Answer: (E)
First note that $3 n+1<2008$ implies that $n<669$.
Since $7 n+1$ is a perfect square, we let $7 n+1=m^{2}$. Then $n=\frac{m^{2}-1}{7}$, which is an integer. Thus 7 is a factor of $m^{2}-1=(m-1)(m+1)$. Since 7 is a prime number, this implies that 7 is a factor of $m-1$ or $m+1$; that is, $m=7 k-1$ or $7 k+1$ for some integer $k$.
If $m=7 k-1$, then $n=\frac{m^{2}-1}{7}=\frac{49 k^{2}-14 k}{7}=7 k^{2}-2 k$. As $n<669$, this gives $7 k^{2}-2 k<669$, which implies that $k=1,2, \ldots, 9$.
Similarly, if $m=7 k+1$, then $n=\frac{m^{2}-1}{7}=\frac{49 k^{2}+14 k}{7}=7 k^{2}+2 k<669$, so we obtain $k=1,2,3, \ldots, 9$.
Hence there are 18 such positive integers.