2005 Alabama ARML TST Problems/Problem 14
Problem
Find the fourth smallest possible value of where x and y are positive integers that satisfy the following equation:
Solution
Solution 1
means that is odd. We can let for some :
y is even, for some .
We need to find all integers such that is twice a perfect square.
Since and are relatively prime, one of them is a perfect square and the other is twice a perfect square. Moreover, the perfect square must be odd.
We will now find four smallest solutions for . Obviously, these will give the four smallest solutions for .
Each time we examine whether the value is a positive integer.
- gives which is not positive.
- gives , hence .
- gives , hence .
- gives .
- gives .
- gives .
- gives .
- gives , hence .
- gives .
- gives .
- gives .
- gives .
- gives .
- gives .
- gives .
- gives .
- gives , hence , and the answer is .
Solution 2
We quickly find the first solution, . Factoring, we get We can square both sides to get So is another solution.
This gives us a way to generate whatever solutions we want to the equation. Raising the first equation to the fourth power gives us The answer is .
See also
2005 Alabama ARML TST (Problems) | ||
Preceded by: Problem 13 |
Followed by: Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |