Difference between revisions of "1973 USAMO Problems/Problem 4"

Revision as of 17:54, 3 July 2013

Problem

Determine all the roots, real or complex, of the system of simultaneous equations

$x+y+z=3$ ,

$x^2+y^2+z^2=3$ ,

$x^3+y^3+z^3=3$ .

Solution

Let $x$ , $y$ , and $z$ be the roots of the cubic polynomial $t^3+at^2+bt+c$ . Let $S_1=x+y+z=3$ , $S_2=x^2+y^2+z^2=3$ , and $S_3=x^3+y^3+z^3=3$ . From this, $S_1+a=0$ , $S_2+aS_1+2b=0$ , and $S_3+aS_2+bS_1+3c=0$ . Solving each of these, $a=-3$ , $b=3$ , and $c=-1$ . Thus $x$ , $y$ , and $z$ are the roots of the polynomial $t^3-3t^2+3t-1=(t-1)^3$ . Thus $x=y=z=1$ , and there are no other solutions.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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 [[Newton's Sums]]
 {{USAMO box|year=1973|num-b=3|num-a=5}}
+{{MAA Notice}}
 [[Category:Olympiad Algebra Problems]]

1973 USAMO (Problems • Resources)
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Difference between revisions of "1973 USAMO Problems/Problem 4"

Revision as of 17:54, 3 July 2013

Problem

Solution

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