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.

Solution 3

Let $a=x-1,$ $b=y-1$ and $c=z-1.$ Then \[a+b+c=0,\] \[a^2+b^2+c^2=0,\] \[a^3+b^3+c^3=0.\] We have \begin{align*} 0&=(a+b+c)^3\\ &=(a^3+b^3+c^3)+3a^2(b+c)+3b^3(a+c)+3c^2(a+b)+6abc\\ &=0-3a^3-3b^3-3c^3+6abc\\ &=6abc. \end{align*} Then one of $a, b$ and $c$ has to be 0, and easy to prove the other two are also 0. So $\boxed{(1,1,1)}$ is the only solution to the system of equations.

J.Z.

Solution 4

We are going to use Intermediate Algebra Techniques to solve this equation.

Let's start with the first one: $x+y+z=3$. This will be referred as the FIRST equation.

We are going to use the first equation to relate to the SECOND one ($x^2+y^2+z^2=3$) and the THIRD one ($x^3+y^3+z^3=3)$.

Squaring this equation: $x^2+y^2+z^2+2xy+2yz+2xz=9$

Subtracting this equation from the 2nd equation in the problem, we have $2xy+2yz+2xz=6$, so $xy+xz+yz=3$.

Now we try the same idea with the cubed terms. Cube the first equation:

$x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3yz^2+3zx^2+3zy^2=27$. Plug in $x^3+y^3+z^3=3$ and factor partially:

$3+3(x^2(y+z)+y^2(x+z)+z^2(x+y))+6xyz=27$.

Now here is the key step. Note that $z=3-x-y, y=3-x-z, x=3-y-z$. So we are going to substitute $y+z, x+z, x+y$ for each of the expressions and we get: $-x^3+3x^2-y^3+3y^2-z^3+3z^2=8-2xyz$ (I rearranged it a bit).

Resubstituting in the second and third equation: $-3+3(3)=8-2xyz$. So $xyz=1$.

So now we have three equations for the elementary symmetric sums of $x,y,z$:

Equation 4: $x+y+z=3$ (this is also equation 1)

Equation 5: $xy+yz+xz=3$

Equation 6: $xyz=1$.

If we call the solutions of $t^3-3t^2+3t-1=0$ (Equation 7) $a,b,c$, then $x,y,z$ are the three roots $a,b,c$ but in some order. Notice that Equation 7 can be factored as $(t-1)^3=0$, which means that $t=1$. Therefore $(x,y,z)$ are permutations of $(1,1,1)$ in some order, which can be only $(1,1,1)$. (This step uses Vieta's formulas)

Therefore, the only solution is $\boxed{(x,y,z)=(1,1,1)}$.

~hastapasta

Solution 5

From the question

Equation 1 : $x+y+z=3$

Equation 2 : $x^2+y^2+z^2=3$

Equation 3 : $x^3+y^3+z^3=3$

From Eq(1) and Eq(2),

Equation 4 : $(x+y+z)^2=9 \implies x^2+y^2+z^2+2\sum xy = 9 \implies \sum xy = 3$

From Eq(1), Eq(3), and the fact that $(x+y+z)^3=x^3+y^3+z^3+3\prod(x+y)$,

Equation 5 : $(x+y+z)^3=x^3+y^3+z^3+3\prod(x+y)= 27 \implies \prod(x+y)=8$

As from Eq(1), $\ x+y=3-z$,$\ y+z=3-x$, $\ z+x = 3- y$

Equation 6 : $\prod(3-x) =8$

Now let there be a cubic function $f(t)$ with the roots $x,y,z$ so from the relation between roots and coefficients,

Equation 7 : $f(t)= t^3-3t^2+3t-d$ (As we don't know the product let the constant be $d$)

As $f(t) = (t-x)(t-y)(t-z)$ then at $t=3$,

Equation 8 : $f(3)=9-d=(3-x)(3-y)(3-z)$

And from Eq(6),

Equation 9 : $d= -1$

So finally,

Equation 10 : $f(t)=t^3-3t^2+3t-1=(t-1)^{3}$

So, $\ (x,y,z)=(1,1,1)$

~Sparrow_1827

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

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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