Difference between revisions of "2000 AIME II Problems/Problem 2"

Latest revision as of 14:51, 21 July 2025

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.

@@ Line 6: / Line 6: @@
 Note that <math>(x-y)</math> and <math>(x+y)</math> have the same [[parity|parities]], so both must be even. We first give a factor of <math>2</math> to both <math>(x-y)</math> and <math>(x+y)</math>. We have <math>2^6 \cdot 5^6</math> left. Since there are <math>7 \cdot 7=49</math> factors of <math>2^6 \cdot 5^6</math>, and since both <math>x</math> and <math>y</math> can be negative, this gives us <math>49\cdot2=\boxed{098}</math> lattice points.
+==Solution 2==
+As with solution 1, note that both <math>x-y</math> and <math>x+y</math> must have the same parities, meaning both have to be even. Additionally, we can express both of them in terms of <math>2^a\cdot3^b</math> and <math>2^c\cdot3^d</math>. Now, <math>a+c</math> 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 <math>b+d</math>, 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 <math>7\cdot7\cdot2</math> which gives <math>098</math>.
+==Solution 3==
+If we restrict ourselves to the first quadrant, this is equivalent to finding [[Pythagorean triple]]s for <math>2000^2 + y^2 = x^2</math>. We know that every Pythagorean triple corresponds to a pair of integers <math>m</math> and <math>n</math> giving:
+<cmath>
+\begin{align*}
+y = m^2 - n^2, && b = 2mn, && x = m^2 + n^2
+\end{align*}
+</cmath>
+If we let <math>b=2mn=2000</math> then each Pythagorean triple corresponds to a factorization <math>mn = 1000 = 2^35^3</math>, of which there are <math>4\times4=25</math>.
+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 <math>y=0</math>, we end up with <math>25\times4-2 = 098</math> 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

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