1977 USAMO Problems/Problem 3: Difference between revisions

Revision as of 08:52, 28 February 2016

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 $p+s(-1-s)-\frac{1}{p}=0$ (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 2: / Line 2: @@
 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}}
-a,b,c,d are roots of equation  <math> x^4+x^3-1=0</math> then by vietas relation
-<cmath>ab +bc+cd+da+ac+bd= 0</cmath>
-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>.
-then sum of roots <math>= ab +bc+cd+da+ac+bd=c/a = -b/a=0</math>
-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
-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>.
-First, <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, abcd = -1</math>.
 Then <math>cd=-\frac{1}{ab}</math> and <math>c+d=-1-(a+b)</math>.
-Remains <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).

1977 USAMO (Problems • Resources)
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