2008 AIME I Problems/Problem 4
Contents
Problem
There exist unique positive integers and that satisfy the equation . Find .
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 .
Solution 6
First complete the square to get . Remember that squares are the sums of consecutive odd integers, so when the difference between the two squares is 244, the two squares must be an even number of odd integers apart. However, there is only one distinct solution, as the problem states, and very quickly you will realize that only two odd integers work. When there are four, then the numbers are not odd, and when it is any other even integer it does not divide. So we need two consecutive odd integers that sum to 244. Easily we find 121 and 123. 121 is the 61st odd integer and 123 is the 62nd odd integer, so is the sum of the first 60 odd integers, or , while is for the same reasons. That way we get , , hence .
-jackshi2006
Video Solution by OmegaLearn
https://youtu.be/euz1azVKUYs?t=424
~ pi_is_3.14
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 |
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