3.48 The population of a city at a certain time is a perfect square. Later, after increasing by 100, the population number is one more than a perfect square. After another increase of 100, the population number is again a perfect square. The original population number is a multiple of which of the following numbers?
(A) 3.
(B) 7.
(C) 9.
(D) 11.
(E) 17.
(13th American High School Mathematics Examination, 1962)
[Solution] Let $N$ be the initial population of the city. According to the problem, we have
$$
N=x^{2}, \quad N+100=y^{2}+1, \quad N+200=z^{2}.
$$

Here $x, y, z$ are all integers. Subtracting the first two equations gives
$$
100=y^{2}-x^{2}+1 \text { or } y^{2}-x^{2}=99 \text {, }
$$

which means $(y+x)(y-x)=99 \cdot 1$ or $33 \cdot 3$ or $11 \cdot 9$.
At this point, $N=49^{2}=2401$, and $N+100=2501=50^{2}+1$, and simultaneously
$$
N+200=2601=51^{2}.
$$

If $\left\{\begin{array}{l}y+x=33, \\ y-x=3 ;\end{array}\right.$ or $\left\{\begin{array}{l}y+x=11, \\ y-x=9 .\end{array}\right.$
the solutions obtained do not satisfy the three equations set up earlier.
Therefore, the answer is $(B)$.