Difference between revisions of "1986 IMO Problems/Problem 1"
(→Solution 3) |
(→Solution 3) |
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<math>13d-1 =z^2 \cdots (3)</math>. | <math>13d-1 =z^2 \cdots (3)</math>. | ||
− | Clearly <math> | + | Clearly <math>z^2+1 = 13d = 3(5d)-2d= 3y^2-x^2+2</math>. |
<math>\implies x^2 +z^2=3y^2+1</math>. | <math>\implies x^2 +z^2=3y^2+1</math>. | ||
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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>z=3r</math> and <math>x= | + | Suppose,<math>z=3r</math> and <math>x=3k</math>±<math>1</math>, |
<math>\implies 3|z </math>, | <math>\implies 3|z </math>, | ||
Revision as of 23:32, 20 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 |