1986 IMO Problems/Problem 1
Contents
Problem
Let be any positive integer not equal to or . Show that one can find distinct in the set such that is not a perfect square.
Solution
Solution 1
We do casework with modular arithmetic.
is not a perfect square.
is not a perfect square.
Therefore, Now consider
is not a perfect square.
is not a perfect square.
As we have covered all possible cases, we are done.
Solution 2
Proof by contradiction:
Suppose , and . From the first equation, is an odd integer. Let . We have , which is an odd integer. Then and must be even integers, denoted by and respectively, and thus , from which can be deduced. Since is even, and have the same parity, so is divisible by . It follows that the odd integer must be divisible by , leading to a contradiction. We are done.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1986 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |