1977 USAMO Problems/Problem 3

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 $x^{6}+x^{4}+x^{3}-x^{2}-1=0$ .

1977 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5
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1977 USAMO Problems/Problem 3

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