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:
![$x^2-2y^2=1$](http://latex.artofproblemsolving.com/b/b/e/bbe557a9f6ab3e6417a1e8771e49d3bdc81ead11.png)
Solution
means that
is odd. We can let
:
y is even, .
We need to find all integers such that
is twice a perfect square.
Since and
are relatively prime, then either
is a perfect square or twice a perfect square, and the same for
.
Let's say that is the perfect square. Then x_1 is odd. We check some:
works.
also works.
We should be able to find some smaller ones when is twice a perfect square, so we try that.
works.
works.
We need to make sure that there is no others below 289. We check all other twice perfect squares, so 289 is the smallest , and then
. Then we get
and
.
.
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 |