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 |