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

(See also)
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==See also==
 
==See also==
 
[[Newton's Sums]]
 
[[Newton's Sums]]
{{USAMO oldbox|year=1973|num-b=3|num-a=5}}
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{{USAMO box|year=1973|num-b=3|num-a=5}}
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Revision as of 14:28, 4 October 2008

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 $x^3+ax^2+bx+c$. 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, $S_1+a=0$, $S_2+aS_1+b=0$, and $S_3+aS_2+bS_1+c=0$. Solving each of these, $a=-3$, $b=3$, and $c=-1$. Thus $x$, $y$, and $z$ are the roots of the polynomial $x^3-3x^2+3x-1=(x-1)^3$. Thus $x+y+z=1$, and there are no other solutions.

See also

Newton's Sums

1973 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All USAMO Problems and Solutions
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