1974 IMO Problems/Problem 3
Prove that the number is not divisible by for any integer
Everything that follows takes place in , i.e. the field we get by adjoining a root of to , the field with elements.
We have . Now, this is zero iff it's zero when we multiply it by , so we may as well prove that . The LHS is from . We have , so by multiplying them we get . If we were to have , then we would get , and this is impossible, since it would make a square in (i.e. would be a quadratic residue modulo , and it's not).
The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: 
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
|1974 IMO (Problems) • Resources|
|1 • 2 • 3 • 4 • 5 • 6||Followed by|
|All IMO Problems and Solutions|