2000 AIME II Problems/Problem 2

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

$(x-y)(x+y)=2000^2=2^8 \cdot 5^6$

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 $49\cdot2=\boxed{098}$ lattice points.

Solution 2

As with solution 1, note that both $x-y$ and $x+y$ must have the same parities, meaning both have to be even. Additionally, we can express both of them in terms of $2^a\cdot3^b$ and $2^c\cdot3^d$ . Now, $a+c$ must be equal to 6, and both have to be greater than or equal to 1, so there are by stars and bars 7 ways to do this. Similarly, for $b+d$ , we have that both only need to be greater than or equal to 0, so this time there are 7 ways to do so. Since both can be negative, we multiply $7\cdot7\cdot2$ which gives $098$ .

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2000 AIME II Problems/Problem 2

Contents

Problem

Solution

Solution 2

See also