1977 USAMO Problems/Problem 3


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


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

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