2005 JBMO Problems/Problem 1
Problem
Find all positive integers satisfying the equation
Solutions
Solution 1
We can re-write the equation as:
or
The above equation tells us that is a perfect square. Since . this implies that
Also, taking on both sides we see that cannot be a multiple of . Also, note that has to be even since is a perfect square. So, cannot be even, implying that is odd.
So we have only to consider for .
Trying above 5 values for we find that result in perfect squares.
Thus, we have cases to check:
Thus all solutions are and .
Solution 2
Expanding, combining terms, and factoring results in Since and are even, must also be even, so and must have the same parity. There are two possible cases.
Case 1: and are both even
Let and . Substitution results in Like before, must be even for the equation to be satisfied. However, if is even, then is a multiple of 4. If and are both even, then is a multiple of 4, but if and are both odd, the is also a multiple of 4. However, is not a multiple of 4, so there are no solutions in this case.
Case 2: and are both odd
Let and , where . Substitution and rearrangement results in Note that , so . There are only a few cases to try out, so we can do guess and check. Rearranging terms once more results in . Since both and are integers, we must have where is an integer. Thus, must be a perfect square.
After trying all values of from 0 to 7, we find that can be or . If , then , and if , then .
Therefore, the ordered pairs that satisfy the original equation are .
See Also
2005 JBMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |