Difference between revisions of "1986 IMO Problems/Problem 1"
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Clearly ,if <math>x^2,z^2</math> is 1 or 0 modulo 3 then it has no solution . | Clearly ,if <math>x^2,z^2</math> is 1 or 0 modulo 3 then it has no solution . | ||
− | Suppose,<math> | + | Suppose,<math>z=3r</math> and <math>x=3</math>±<math>1</math>, |
<math>\implies 3|z </math>, | <math>\implies 3|z </math>, | ||
Revision as of 00:32, 21 January 2020
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 mods.
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.
Solution 3
Suppose one can't find distinct a,b from the set such that
is a perfect square.
Let,
.
Clearly .
.
Clearly ,if is 1 or 0 modulo 3 then it has no solution .
Suppose, and
±
,
,
.
So, and
.
.
It is contradiction ! Since .
~ @ftheftics
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 |