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Difference between revisions of "1971 Canadian MO Problems/Problem 5"

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=== Note ===
 
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You can arrive at the same conclusion by observing that multiplying 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 <math>2</math>, odd times odd = odd because no new factors of <math>2</math> are introduced. (You can also use the <math>2n</math> and <math>2k+1</math> method.) The given of the solution, again, implies <math>a_n + a_{n-1} + \dots + a_1</math> is even. Since the parity of none of the terms change if you multiply the coefficients <math>a_i</math> by <math>m^i</math>, the sum of all of the terms stays even. Adding <math>a_0</math> will make this sum odd, and since <math>0</math> is an even number, this is a contradiction.
+
You can arrive at the same conclusion by observing that multiplying by an odd number does not change the parity (odd/even) of a number.  
 +
 
 +
 
 +
The proof of this is simple: All even numbers have a factor of <math>2</math>. Odd times even = even because odd numbers do not take away a factor of <math>2</math>, odd times odd = odd because no new factors of <math>2</math> are introduced. (You can also use the <math>2n</math> and <math>2k+1</math> method.)
 +
 
 +
 
 +
The given of the solution, again, implies <math>a_n + a_{n-1} + \dots + a_1</math> is even. Since the parity of none of the terms change if you multiply the coefficients <math>a_i</math> by <math>m^i</math>, the sum of all of the terms stays even. Adding <math>a_0</math> to <math>a_n m^n + a_{n-1} m^{n-1} + \dots + a_1 m^1</math> will make this sum odd, and since <math>0</math> is an even number, this is a contradiction.
  
 
~jasminelover7254
 
~jasminelover7254

Latest revision as of 18:55, 20 December 2024

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 multiplying by an odd number does not change the parity (odd/even) of a number.


The proof of this is simple: All even numbers have a factor of $2$. 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.)


The given of the solution, again, implies $a_n + a_{n-1} + \dots + a_1$ is even. Since the parity of none of the terms change if you multiply the coefficients $a_i$ by $m^i$, the sum of all of the terms stays even. Adding $a_0$ to $a_n m^n + a_{n-1} m^{n-1} + \dots + a_1 m^1$ 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
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