math-algebra-01

# Power Sums of a Symmetric System

Real numbers $x$, $y$, $z$ satisfy the system

$$
\begin{aligned}
x + y + z &= 6,\\
x^2 + y^2 + z^2 &= 14,\\
x^3 + y^3 + z^3 &= 36.
\end{aligned}
$$

Find the value of $x^4 + y^4 + z^4$.

Give your final answer as a single integer.

Let $S_k = x^k + y^k + z^k$. We are given:
$$S_1 = 6$$
$$S_2 = 14$$
$$S_3 = 36$$
We want to find $S_4$.

Let $e_1, e_2, e_3$ be the elementary symmetric polynomials in $x, y, z$:
$$e_1 = x + y + z$$
$$e_2 = xy + yz + zx$$
$$e_3 = xyz$$

From the given information, we have:
$$e_1 = S_1 = 6$$

We can find $e_2$ using the identity $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx)$, which translates to $e_1^2 = S_2 + 2e_2$.
$$6^2 = 14 + 2e_2$$
$$36 = 14 + 2e_2$$
$$22 = 2e_2$$
$$e_2 = 11$$

Next, we find $e_3$. We can use the identity for the sum of cubes:
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - (xy+yz+zx))$$
In terms of our variables:
$$S_3 - 3e_3 = e_1(S_2 - e_2)$$
Substituting the known values:
$$36 - 3e_3 = 6(14 - 11)$$
$$36 - 3e_3 = 6(3)$$
$$36 - 3e_3 = 18$$
$$-3e_3 = 18 - 36$$
$$-3e_3 = -18$$
$$e_3 = 6$$

Now we know the elementary symmetric polynomials are $e_1 = 6$, $e_2 = 11$, and $e_3 = 6$. The numbers $x, y, z$ are the roots of the cubic polynomial:
$$P(t) = t^3 - e_1 t^2 + e_2 t - e_3 = 0$$
$$t^3 - 6t^2 + 11t - 6 = 0$$

Since $x, y, z$ are roots of this equation, they satisfy:
$$x^3 - 6x^2 + 11x - 6 = 0$$
$$y^3 - 6y^2 + 11y - 6 = 0$$
$$z^3 - 6z^2 + 11z - 6 = 0$$

To find $S_4 = x^4 + y^4 + z^4$, we multiply the equation for $x$ by $x$, the equation for $y$ by $y$, and the equation for $z$ by $z$:
$$x^4 - 6x^3 + 11x^2 - 6x = 0$$
$$y^4 - 6y^3 + 11y^2 - 6y = 0$$
$$z^4 - 6z^3 + 11z^2 - 6z = 0$$

Summing these three equations gives:
$$(x^4 + y^4 + z^4) - 6(x^3 + y^3 + z^3) + 11(x^2 + y^2 + z^2) - 6(x + y + z) = 0$$
$$S_4 - 6S_3 + 11S_2 - 6S_1 = 0$$

Now we substitute the known values of $S_1, S_2, S_3$:
$$S_4 - 6(36) + 11(14) - 6(6) = 0$$
$$S_4 - 216 + 154 - 36 = 0$$
$$S_4 - 252 + 154 = 0$$
$$S_4 - 98 = 0$$
$$S_4 = 98$$

Alternatively, we could solve the cubic equation $t^3 - 6t^2 + 11t - 6 = 0$. By inspection, $t=1$ is a root ($1-6+11-6=0$). Dividing by $(t-1)$, we get $(t-1)(t^2-5t+6)=0$, which factors further to $(t-1)(t-2)(t-3)=0$. Thus, the roots are $\{1, 2, 3\}$.
Checking the sums:
$1+2+3=6$
$1^2+2^2+3^2 = 1+4+9=14$
$1^3+2^3+3^3 = 1+8+27=36$
Then $S_4 = 1^4 + 2^4 + 3^4 = 1 + 16 + 81 = 98$.

\boxed{98}