# 1973 USAMO Problems/Problem 4

## 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.

## Solution 2

Let $P(t)=t^3-at^2+bt-c$ have roots x, y, and z. Then $\[0=P(x)+P(y)+P(z)=3-3a+3b-3c\]$ using our system of equations, so $P(1)=0$. Thus, at least one of x, y, and z is equal to 1; without loss of generality, let $x=1$. Then we can use the system of equations to find that $y=z=1$ as well, and so $\boxed{(1,1,1)}$ is the only solution to the system of equations.

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

## See Also

 1973 USAMO (Problems • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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