1971 Canadian MO Problems/Problem 5
Problem
Let , where the coefficients are integers. If and are both odd, show that has no integral roots.
Solution
Inputting and into , we obtain
and
The problem statement tells us that these are both odd. We will keep this in mind as we begin our proof by contradiction.
Suppose for the sake of contradiction that there exist integer such that
Substitution gives
By the Integer Root Theorem, must divide . Since is odd, as shown above, must be odd. We also know that must be even since it is equal to . From above, we have that must be odd. Since we also have that is odd, must be even. Thus, there must be an even number of odd for integer . Thus, the sum of all must be even. Then for all that are even for integer we must have the sum of all even since every is even. In conclusion, we have
even. But since is odd, must be odd. Thus, it cannot equal and we have arrived at a contradiction.
-Solution by thecmd999
See Also
1971 Canadian MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |