Difference between revisions of "1986 IMO Problems/Problem 1"
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{{IMO box|year=1986|before=First Problem|num-a=2}} | {{IMO box|year=1986|before=First Problem|num-a=2}} |
Latest revision as of 15:02, 31 July 2022
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 |