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

Revision as of 21:22, 20 September 2022

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

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 $p+s(-1-s)-\frac{1}{p}=0$ \hskip 3pt (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 $x^{6}+x^{4}+x^{3}-x^{2}-1=0$ .

@@ Line 11: / Line 11: @@
 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>, so <math>p+s(-1-s)-\frac{1}{p}=0</math> \hskip 3pt (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>.

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