Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1971 Canadian MO Problems/Problem 5

Revision as of 18:35, 20 December 2024 by Wes (talk | contribs) (→‎Solution)

Problem

Let $p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x+a_0$, where the coefficients $a_i$ are integers. If $p(0)$ and $p(1)$ are both odd, show that $p(x)$ has no integral roots.

Solution

Inputting $0$ and $1$ into $p(x)$, we obtain

$p(0)=a_0$

and

$p(1)=a_0+a_1+a_2+\cdots+a_n$

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 $m$ such that

$p(m)=0$

Substitution gives

$a_nm^n + a_{n-1}m^{n-1} + \cdots + a_1m+a_0=0$

By the Integer Root Theorem, $m$ must divide $a_0$. Since $a_0$ is odd, as shown above, $m$ must be odd. We also know that $p(m)$ must be even since it is equal to $0$. From above, we have that $a_0+a_1+a_2+\cdots+a_n$ must be odd. Since we also have that $a_0$ is odd, $a_1+a_2+a_3+\cdots+a_n$ must be even. Thus, there must be an even number of odd $a_i$ for integer $0<i<n+1$. Thus, the sum of all $a_im^i$ must be even. Then for all $a_k$ that are even for integer $0<k<n+1$ we must have the sum of all $a_km^k$ even since every $a_km^k$ is even. In conclusion, we have

$a_nm^n + a_{n-1}m^{n-1} + \cdots + a_1m$

even. But since $a_0$ is odd, $p(m)$ must be odd. Thus, it cannot equal $0$ and we have arrived at a contradiction. $Q.E.D.$

-Solution by thecmd999

Note

You can arrive at the same conclusion by observing that multiply by an odd number does not change the parity (odd/even) of a number. The proof of this is simple: odd times even = even because odd numbers do not take away a factor of $2$, odd times odd = odd because no new factors of $2$ are introduced. (You can also use the $2n$ and $2k+1$ method.) Since the parity of none of the coefficients change if you multiply by $m^k$, the sum of all terms are even. Adding $a_0$ will make this sum odd, and since $0$ is an even number, this is a contradiction. ~jasminelover7254

See Also

1971 Canadian MO (Problems)
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 7
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1971_Canadian_MO_Problems/Problem_5&oldid=237429"