Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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 $098$.

Solution 3

If we restrict ourselves to the first quadrant, this is equivalent to finding Pythagorean triples for $2000^2 + y^2 = x^2$. We know that every Pythagorean triple corresponds to a pair of integers $m$ and $n$ giving:

\begin{align*} y = m^2 - n^2, && b = 2mn, && x = m^2 + n^2 \end{align*}

If we let $b=2mn=2000$ then each Pythagorean triple corresponds to a factorization $mn = 1000 = 2^35^3$, of which there are $4\times4=25$.

But we've been only looking at the first quadrant. If we reflect this quadrant to the others, and eliminate the two duplicate reflections where $y=0$, we end up with $25\times4-2 = 098$ solutions.

See also

2000 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_2&oldid=254922"