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

(Created page with "<math>p(0)=a_nx^{n}+a_{n-1}x^{n-1}+...a_1x+a_0=a_0</math> We know that p(0) is odd, so we know <math>a_0</math> is odd. By Vieta's this means that all the roots of polynomial p...")
 
(→‎See Also)
 
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<math>p(0)=a_nx^{n}+a_{n-1}x^{n-1}+...a_1x+a_0=a_0</math>
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== Problem ==
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Let <math>p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x+a_0</math>, where the coefficients <math> a_i</math> are integers. If <math>p(0)</math> and <math>p(1)</math> are both odd, show that <math>p(x)</math> has no integral roots.
  
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We know that p(0) is odd, so we know <math>a_0</math> is odd.
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== Solution ==
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Inputting <math>0</math> and <math>1</math> into <math>p(x)</math>, we obtain
  
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By Vieta's this means that all the roots of polynomial p(x) are also odd.
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<math>p(0)=a_0</math>
  
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<math>p(1)=a_n(x-r_1)(x-r_2)...(x-r_n)=a_n(1-r_1)(1-r_2)...(1-r_n)</math>
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and
  
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Since all the roots <math>r_k</math> where <math>{1}\le{k}\le{n}</math> are odd, <math>1-r_k</math> must be even, and <math>p(1)</math> must be even.
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<math>p(1)=a_0+a_1+a_2+\cdots+a_n</math>
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The problem statement tells us that these are both odd. We will keep this in mind as we begin our proof by contradiction.
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Suppose for the sake of contradiction that there exist integer <math>m</math> such that
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<math>p(m)=0</math>
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Substitution gives
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<math>a_nm^n + a_{n-1}m^{n-1} + \cdots + a_1m+a_0=0</math>
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By the Integer Root Theorem, <math>m</math> must divide <math>a_0</math>. Since <math>a_0</math> is odd, as shown above, <math>m</math> must be odd. We also know that <math>p(m)</math> must be even since it is equal to <math>0</math>. From above, we have that <math>a_0+a_1+a_2+\cdots+a_n</math> must be odd. Since we also have that <math>a_0</math> is odd, <math>a_1+a_2+a_3+\cdots+a_n</math> must be even. Thus, there must be an even number of odd <math>a_i</math> for integer <math>0<i<n+1</math>. Thus, the sum of all <math>a_im^i</math> must be even. Then for all <math>a_k</math> that are even for integer <math>0<k<n+1</math> we must have the sum of all <math>a_km^k</math> even since every <math>a_km^k</math> is even. In conclusion, we have
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<math>a_nm^n + a_{n-1}m^{n-1} + \cdots + a_1m</math>
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even. But since <math>a_0</math> is odd, <math>p(m)</math> must be odd. Thus, it cannot equal <math>0</math> and we have arrived at a contradiction. <math>Q.E.D.</math>
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-Solution by '''thecmd999'''
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== See Also ==
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{{Old CanadaMO box|num-b=5|num-a=7|year=1971}}
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[[Category:Olympiad Algebra Problems]]

Latest revision as of 18:57, 7 August 2016

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

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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