The number of integral points on the hyperbola $x^2 - y^2 = (2000)^2$ is ____? (An integral point is a point both of whose coordinates are integers.
My attempt: The equation can be rewritten as: $$(x-y)(x+y) = 2000^2 = 2^8 \cdot 5^6$$
Now, both of $x+y$ and $x-y$ need to be odd or even. It is impossible for both of them to be odd, hence both of them must be even. The number of even pairs of factors of $2^8 \cdot 5^6$ is $7 \times 7 = 49$. Hence, the total number of integral points is $\boxed{49}$
The answer: The textbook claims the answer to be $\boxed{98}$ and counts the number of even pairs of factors as $7 \times 7 \times 2 = 98$. Could someone point out where I am going wrong? (or if the textbook has made an error)
EDIT: This question has been answered but I wanted to add that the 2000 seemed conspicuous, so I went to Approach0. This problem is taken from AIME 2000. the referenced AoPS link also does a great job explaining the solution.
The number of even factors of $2^8\cdot 5^6$ is $8\cdot 7=56$. You can have $1$ to $8$ factors of $2$ and $0$ to $6$ factors of $5$. You need both factors to be even, so the first one can have from $1$ to $7$ factors of $2$, which gives $49$ factorizations into two even numbers.
We insist that $|x| \ge |y|$ so that the difference of squares is positive. The $49$ factorizations include $24$ with the absolute values of factors different, each in two orders, plus the one factorization $2000 \cdot 2000$. Each of the $24$ that have $|x| \gt |y|$ give four choices of signs, while $2000^2-0^2$ gives two choices of signs, for a total of $98$ solutions.