Difference between revisions of "2008 AIME I Problems/Problem 4"
Mathgeek2006 (talk | contribs) m (→Solution 2) |
Mathgeek2006 (talk | contribs) m (→Solution 2) |
||
Line 14: | Line 14: | ||
We complete the square like in the first solution: <math>y^2 = (x+42)^2 + 244</math>. Since consecutive squares differ by the consecutive odd numbers, we note that <math>y</math> and <math>x+42</math> must differ by an even number. We can use casework with the even numbers, starting with <math>y-(x+42)=2</math>. | We complete the square like in the first solution: <math>y^2 = (x+42)^2 + 244</math>. Since consecutive squares differ by the consecutive odd numbers, we note that <math>y</math> and <math>x+42</math> must differ by an even number. We can use casework with the even numbers, starting with <math>y-(x+42)=2</math>. | ||
<cmath>\begin{align*}2(x+42)+1+2(x+42)+3&=244\\ | <cmath>\begin{align*}2(x+42)+1+2(x+42)+3&=244\\ | ||
− | \ | + | \Leftrightarrow x&=18\end{align*}</cmath> |
Thus, <math>y=62</math> and the answer is <math>\boxed{080}</math>. | Thus, <math>y=62</math> and the answer is <math>\boxed{080}</math>. |
Revision as of 17:27, 13 March 2015
Problem
There exist unique positive integers and that satisfy the equation . Find .
Contents
Solution
Solution 1
Completing the square, . Thus by difference of squares.
Since is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since , the factors must be and . Since , we have and ; the latter equation implies that .
Indeed, by solving, we find is the unique solution.
Solution 2
We complete the square like in the first solution: . Since consecutive squares differ by the consecutive odd numbers, we note that and must differ by an even number. We can use casework with the even numbers, starting with .
Thus, and the answer is .
Solution 3
We see that . By quadratic residues, we find that either . Also, , so . Combining, we see that .
Testing and other multiples of , we quickly find that is the solution.
Solution 4
We solve for x:
So is a perfect square. Since 244 is even, the difference is even, so we try : , .
Plugging into our equation, we find that , and indeed satisfies the original equation.
Solution 5
Let for some , substitute into the original equation to get .
All terms except for the last one are even, hence must be even, hence let . We obtain . Rearrange to .
Obviously for the right hand side is negative and the left hand side is positive. Hence . Let , then .
We have . Left hand side simplifies to . As must be an integer, must divide the left hand side. But is a prime, which only leaves two options: and .
Option gives us a negative . Option gives us , and , hence .
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.