Difference between revisions of "1993 IMO Problems/Problem 1"
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+ | It’s actually trivial by Perron’s Criterion lol | ||
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Note: Quoting Perron's Criterion on the actual IMO will very likely result in a score in the set <math>\{0,1\}</math>, since it was not a well-known result back then. | Note: Quoting Perron's Criterion on the actual IMO will very likely result in a score in the set <math>\{0,1\}</math>, since it was not a well-known result back then. |
Latest revision as of 06:08, 22 June 2024
Contents
[hide]Problem
Let , where is an integer. Prove that cannot be expressed as the product of two non-constant polynomials with integer coefficients.
Solution
For the sake of contradiction, assume that for polynomials and in . Furthermore, let with if and with if . This gives that .
We have that , or . WLOG, let (and thus ). Since and divides but not , we need that . We can keep on going up the chain until we get that . Then, by equating coefficients once more, we get that . Taking the equation gives that . This implies that . Thus, the degree of is at least . However, because is a non-constant factor of , we have that the degree of is at most . Thus, the degree of is , implying that the degree of is .
From this fact, we have that there must exist an integer root of . However, when is an integer, , meaning is odd, so cannot be a root of .
Hence, cannot be expressed as for polynomials and in . This means that cannot be expressed as the product of two non-constant polynomials with integer coefficients.
Q.E.D.
Alternate Solution
It’s actually trivial by Perron’s Criterion lol
Note: Quoting Perron's Criterion on the actual IMO will very likely result in a score in the set , since it was not a well-known result back then.
See Also
1993 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |