A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 - y^2 = 2000^2$?

\[(x-y)(x+y)=2000^2=2^8 \cdot 5^6\]
Note that $(x-y)$ and $(x+y)$ have the same parities, so both must be even. We first give a factor of $2$ to both $(x-y)$ and $(x+y)$. We have $2^6 \cdot 5^6$ left. Since there are $7 \cdot 7=49$ factors of $2^6 \cdot 5^6$, and since both $x$ and $y$ can be negative, this gives us $49\cdot2=\boxed{98}$ lattice points.
98
Level 5
Number Theory
Problem: A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 - y^2 = 2000^2$?

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the lattice points on the hyperbola defined by the equation \(x^2 - y^2 = 2000^2\), we start by rewriting the equation in a more manageable form. This can be factored as \((x - y)(x + y) = 2000^2\). 

Next, we recognize that \(2000^2\) can be expressed in terms of its prime factorization. The prime factorization of \(2000\) is \(2^4 \times 5^3\), so \(2000^2 = 2^8 \times 5^6\). 

The number of factors of \(2000^2\) can be calculated using the formula for the number of divisors based on prime factorization. If \(n = p_1^{e_1} \times p_2^{e_2} \times \ldots \times p_k^{e_k}\), then the number of divisors \(d(n)\) is given by \((e_1 + 1)(e_2 + 1) \ldots (e_k + 1)\). 

Applying this to \(2000^2\):
- For \(2^8\), the exponent is \(8\), so \(8 + 1 = 9\).
- For \(5^6\), the exponent is \(6\), so \(6 + 1 = 7\).

Thus, the total number of divisors of \(2000^2\) is \(9 \times 7 = 63\).

Each divisor \(d\) of \(2000^2\) corresponds to a pair of integers \((x - y, x + y)\) such that \(x - y = d\) and \(x + y = \frac{2000^2}{d}\). From these equations, we can solve for \(x\) and \(y\):
1. \(x = \frac{(x - y) + (x + y)}{2} = \frac{d + \frac{2000^2}{d}}{2}\)
2. \(y = \frac{(x + y) - (x - y)}{2} = \frac{\frac{2000^2}{d} - d}{2}\)

For \(x\) and \(y\) to be integers, both \(d + \frac{2000^2}{d}\) and \(\frac{2000^2}{d} - d\) must be even. This occurs when both \(d\) and \(\frac{2000^2}{d}\) are either both even or both odd. 

Since \(2000^2\) is even, all divisors \(d\) must also be even. The even divisors of \(2000^2\) can be found by considering the prime factorization. The even divisors are derived from the factors of \(2^8\) and \(5^6\) where the exponent of \(2\) is at least \(1\). 

The number of even divisors can be calculated as follows:
- For \(2^8\), the possible exponents are \(1\) through \(8\) (which gives \(8\) choices).
- For \(5^6\), the possible exponents are \(0\) through \(6\) (which gives \(7\) choices).

Thus, the total number of even divisors is \(8 \times 7 = 56\).

However, each pair \((d, \frac{2000^2}{d})\) gives rise to two lattice points \((x, y)\) and \((x, -y)\), except when \(y = 0\). We need to check if \(y = 0\) is possible. Setting \(y = 0\) in the original equation gives \(x^2 = 2000^2\), leading to \(x = 2000\) or \(x = -2000\), which are two additional lattice points.

Thus, the total number of lattice points is \(2 \times 56 + 2 = 112\). However, we must account for the fact that we only want unique lattice points. Since the hyperbola is symmetric about both axes, we need to divide the count of unique points by \(2\) to avoid double counting.

Therefore, the final count of unique lattice points on the hyperbola is \(98\).<|END_OF_LATENT|> The final answer is $98$.