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1977 USAMO Problems/Problem 3: Difference between revisions

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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>.


First, Vieta's relations give <math>a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abcd = -1</math>.
First, Vieta's relations give <math>a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abc+abd+acd+bcd=0, abcd = -1</math>.


Then <math>cd=-\frac{1}{ab}</math> and <math>c+d=-1-(a+b)</math>.
Then <math>cd=-\frac{1}{ab}</math> and <math>c+d=-1-(a+b)</math>.
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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>.
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).
Let <math>a+b=s</math> and <math>ab=p</math>.


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>.
Thus, <math>0=ab+ac+ad+bc+bd+cd=p+s(-1-s)-\frac{1}{p}</math>. &nbsp; (1)


Multiplying: <math>a^{3}b^{3}(a+1)(b+1)=1</math> or <math>p^{3}(p+s+1)=1</math>.
Also, <math>0=abc+abd+acd+bcd=p(-1-s)-s/p</math>.


Solving <math>s= \frac{1-p^{4}-p^{3}}{p^{3}}</math>.
Solving this equation for <math>s</math>, <math>s= \frac{-p^2}{p^2+1}</math>.


In (1): <math>\frac{p^{8}+p^{5}-2p^{4}-p^{3}+1}{p^{6}}=0</math>.
Substituting into (1): <math>\frac{p^{6}+p^{4}+p^{3}-p^{2}-1}{p(p^2+1)^2}=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>.
Conclusion: <math>p =ab</math> is a root of <math>x^{6}+x^{4}+x^{3}-x^{2}-1=0</math>.

Latest revision as of 21:46, 20 September 2022

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 $a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abc+abd+acd+bcd=0, abcd = -1$.

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 $ab=p$.

Thus, $0=ab+ac+ad+bc+bd+cd=p+s(-1-s)-\frac{1}{p}$.   (1)

Also, $0=abc+abd+acd+bcd=p(-1-s)-s/p$.

Solving this equation for $s$, $s= \frac{-p^2}{p^2+1}$.

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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