Difference between revisions of "1977 USAMO Problems/Problem 3"

m
(Solution)
(5 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
If <math> a</math> and <math> b</math> are two of the roots of <math> x^4\plus{}x^3\minus{}1\equal{}0</math>, prove that <math> ab</math> is a root of <math> x^6\plus{}x^4\plus{}x^3\minus{}x^2\minus{}1\equal{}0</math>.
+
If <math> a</math> and <math> b</math> are two of the roots of <math> x^4+x^3-1=0</math>, prove that <math> ab</math> is a root of <math> x^6+x^4+x^3-x^2-1=0</math>.
  
== Solution ==
+
==Solution==
{{solution}}
+
Given the roots <math>a,b,c,d</math> of the equation <math>x^{4}+x^{3}-1=0</math>.
 +
 
 +
First, Vieta's relations give <math>a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abcd = -1</math>.
 +
 
 +
Then <math>cd=-\frac{1}{ab}</math> and <math>c+d=-1-(a+b)</math>.
 +
 
 +
The other coefficients give <math>ab+(a+b)(c+d)+cd = 0</math> or <math>ab+(a+b)[-1-(a+b)]-\frac{1}{ab}=0</math>.
 +
 
 +
Let <math>a+b=s</math> and <math>ab=p</math>, so <math>p+s(-1-s)-\frac{1}{p}=0</math>(1).
 +
 
 +
Second, <math>a</math> is a root, <math>a^{4}+a^{3}=1</math> and <math>b</math> is a root, <math>b^{4}+b^{3}=1</math>.
 +
 
 +
Multiplying: <math>a^{3}b^{3}(a+1)(b+1)=1</math> or <math>p^{3}(p+s+1)=1</math>.
 +
 
 +
Solving <math>s= \frac{1-p^{4}-p^{3}}{p^{3}}</math>.
 +
 
 +
In (1): <math>\frac{p^{8}+p^{5}-2p^{4}-p^{3}+1}{p^{6}}=0</math>.
 +
 
 +
<math>p^{8}+p^{5}-2p^{4}-p^{3}+1=0</math> or <math>(p-1)(p+1)(p^{6}+p^{4}+p^{3}-p^{2}-1)= 0</math>.
 +
 
 +
Conclusion: <math>p =ab</math> is a root of <math>x^{6}+x^{4}+x^{3}-x^{2}-1=0</math>.
  
 
== See Also ==
 
== See Also ==
 
{{USAMO box|year=1977|num-b=2|num-a=4}}
 
{{USAMO box|year=1977|num-b=2|num-a=4}}
 +
{{MAA Notice}}
  
[[Category:Olympiad AlgebraProblems]]
+
[[Category:Olympiad Algebra Problems]]

Revision as of 09:52, 28 February 2016

Problem

If $a$ and $b$ are two of the roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$.

Solution

Given the roots $a,b,c,d$ of the equation $x^{4}+x^{3}-1=0$.

First, Vieta's relations give $a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abcd = -1$.

Then $cd=-\frac{1}{ab}$ and $c+d=-1-(a+b)$.

The other coefficients give $ab+(a+b)(c+d)+cd = 0$ or $ab+(a+b)[-1-(a+b)]-\frac{1}{ab}=0$.

Let $a+b=s$ and $ab=p$, so $p+s(-1-s)-\frac{1}{p}=0$(1).

Second, $a$ is a root, $a^{4}+a^{3}=1$ and $b$ is a root, $b^{4}+b^{3}=1$.

Multiplying: $a^{3}b^{3}(a+1)(b+1)=1$ or $p^{3}(p+s+1)=1$.

Solving $s= \frac{1-p^{4}-p^{3}}{p^{3}}$.

In (1): $\frac{p^{8}+p^{5}-2p^{4}-p^{3}+1}{p^{6}}=0$.

$p^{8}+p^{5}-2p^{4}-p^{3}+1=0$ or $(p-1)(p+1)(p^{6}+p^{4}+p^{3}-p^{2}-1)= 0$.

Conclusion: $p =ab$ is a root of $x^{6}+x^{4}+x^{3}-x^{2}-1=0$.

See Also

1977 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png