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Difference between revisions of "1977 USAMO Problems/Problem 3"
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== Problem ==
 
== Problem ==
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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>.
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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}}
 
{{solution}}
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a,b,c,d are roots of equation  <math> x^4\plus{}x^3\minus{}1\equal{}0</math> then by vietas relation
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a,b,c,d are roots of equation  <math> x^4+x^3-1=0</math> then by vietas relation
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ab +bc+cd+da+ac+bd=c/a = 0
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<cmath>ab +bc+cd+da+ac+bd= 0</cmath>
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let us suppose ab,bc,cd,da,ac,bd are roots of <math> x^6\plus{}x^4\plus{}x^3\minus{}x^2\minus{}1\equal{}0</math>.
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let us suppose <math>ab,bc,cd,da,ac,bd</math> are roots of <math> x^6+x^4+x^3-x^2-1=0</math>.
  
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then sum of roots = ab +bc+cd+da+ac+bd=c/a = -b/a=0
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then sum of roots <math>= ab +bc+cd+da+ac+bd=c/a = -b/a=0</math>
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sum taken two at a time= abxbc + bcxca +..........=c/a=1
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sum taken two at a time <math>= ab\times bc + bc\times ca +..........=c/a=1</math>
 
similarly we prove for the roots taken three four five and six at a time
 
similarly we prove for the roots taken three four five and six at a time
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to prove ab,bc,cd,da,ac,bd are roots of second equation
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to prove <math>ab,bc,cd,da,ac,bd</math> are roots of second equation
  
 
Given the roots <math>a,b,c,d</math> of the equation <math>x^{4}+x^{3}-1=0</math>.
 
Given the roots <math>a,b,c,d</math> of the equation <math>x^{4}+x^{3}-1=0</math>.

Revision as of 22:05, 13 March 2015

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

This problem needs a solution. If you have a solution for it, please help us out by adding it. a,b,c,d are roots of equation $x^4+x^3-1=0$ then by vietas relation \[ab +bc+cd+da+ac+bd= 0\] let us suppose $ab,bc,cd,da,ac,bd$ are roots of $x^6+x^4+x^3-x^2-1=0$.

then sum of roots $= ab +bc+cd+da+ac+bd=c/a = -b/a=0$ sum taken two at a time $= ab\times bc + bc\times ca +..........=c/a=1$ similarly we prove for the roots taken three four five and six at a time to prove $ab,bc,cd,da,ac,bd$ are roots of second equation

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

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

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

Remains $ab+(a+b)(c+d)+cd = 0$ 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 (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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