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...")
 
(I LEAVE THIS PROBLEM OPEN. COME SOLVE IT.)
Line 1: Line 1:
<math>p(0)=a_nx^{n}+a_{n-1}x^{n-1}+...a_1x+a_0=a_0</math>
+
== Problem ==
 +
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.
  
We know that p(0) is odd, so we know <math>a_0</math> is odd.
+
== Solution ==
 +
{{solution}}
  
By Vieta's this means that all the roots of polynomial p(x) are also odd.
+
== See Also ==
  
<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>
+
[[Category:Intermediate Algebra Problems]]
 
 
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.
 

Revision as of 15:52, 12 September 2012

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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See Also