6. Find the number of ordered pairs $(x, y)$, where $x$ is an integer and $y$ is a perfect square, such that $y=(x-90)^{2}-4907$.
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
6. Answer: (E)
Let $y=m^{2}$ and $(x-90)^{2}=k^{2}$, where $m$ and $k$ are positive integers. Then we obtain $k^{2}-m^{2}=4907=7 \times 701=1 \times 4907$, which gives
$$
(k-m)(k+m)=7 \times 701 \text { or }(k-m)(k+m)=1 \times 4907 .
$$

It follows that
$k-m=7$ and $k+m=701$, or $k-m=1$ and $k+m=4907$.

Solving these two pairs of equations gives
$$
(k, m)=(354,347) \text { and }(k, m)=(2454,2453) \text {. }
$$

Therefore the ordered pairs $(x, y)$ that satisfy the given equation are:
$$
\left(444,347^{2}\right),\left(-264,347^{2}\right),\left(2544,2453^{2}\right),\left(-2364,2453^{2}\right) \text {. }
$$

Hence the answer is 4 .