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:
$S_1 = x + y + z = 6$
$S_2 = x^2 + y^2 + z^2 = 14$
$S_3 = x^3 + y^3 + z^3 = 36$

We want to find $S_4 = x^4 + y^4 + z^4$.

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

We know the relations between power sums and elementary symmetric polynomials (Newton sums).
For $n=1, 2, 3, \dots$:
$S_1 - e_1 = 0$
$S_2 - e_1 S_1 + 2e_2 = 0$
$S_3 - e_1 S_2 + e_2 S_1 - 3e_3 = 0$
$S_4 - e_1 S_3 + e_2 S_2 - e_3 S_1 = 0$

Let's calculate $e_1, e_2, e_3$ from the given values.

From $S_1 = 6$, we have:
$e_1 = 6$.

From $S_2 = 14$, we use the relation:
$S_2 - e_1 S_1 + 2e_2 = 0$
$14 - 6(6) + 2e_2 = 0$
$14 - 36 + 2e_2 = 0$
$-22 + 2e_2 = 0$
$2e_2 = 22$
$e_2 = 11$.

From $S_3 = 36$, we use the relation:
$S_3 - e_1 S_2 + e_2 S_1 - 3e_3 = 0$
$36 - 6(14) + 11(6) - 3e_3 = 0$
$36 - 84 + 66 - 3e_3 = 0$
Let's compute the sum:
$36 + 66 = 102$
$102 - 84 = 18$
So, $18 - 3e_3 = 0$
$3e_3 = 18$
$e_3 = 6$.

So the elementary symmetric polynomials are:
$e_1 = 6$
$e_2 = 11$
$e_3 = 6$

The numbers $x, y, z$ are the roots of the cubic polynomial:
$P(t) = t^3 - e_1 t^2 + e_2 t - e_3 = 0$
$P(t) = t^3 - 6t^2 + 11t - 6 = 0$

Let's check if this polynomial has real roots.
We can try to factor it.
Possible rational roots are divisors of 6: $\pm 1, \pm 2, \pm 3, \pm 6$.
Let's test $t=1$:
$1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$.
So $t=1$ is a root.
We can divide $P(t)$ by $(t-1)$.
$t^3 - 6t^2 + 11t - 6 = (t-1)(t^2 - 5t + 6)$.
Now factor $t^2 - 5t + 6$.
$(t-2)(t-3) = t^2 - 5t + 6$.
So the roots are $1, 2, 3$.
Thus, the set $\{x, y, z\}$ is $\{1, 2, 3\}$.
Let's check the given conditions with these values.
$x+y+z = 1+2+3 = 6$. Correct.
$x^2+y^2+z^2 = 1^2+2^2+3^2 = 1+4+9 = 14$. Correct.
$x^3+y^3+z^3 = 1^3+2^3+3^3 = 1+8+27 = 36$. Correct.

Now we need to find $S_4 = x^4 + y^4 + z^4$.
Using the roots $1, 2, 3$:
$S_4 = 1^4 + 2^4 + 3^4$
$S_4 = 1 + 16 + 81$
$S_4 = 98$.

Alternatively, we can use Newton sums without finding the roots explicitly.
We have the recurrence relation for $S_k$:
$S_k - e_1 S_{k-1} + e_2 S_{k-2} - e_3 S_{k-3} = 0$ for $k \ge 3$.
Wait, the general formula is:
$S_k - e_1 S_{k-1} + e_2 S_{k-2} - e_3 S_{k-3} = 0$
Let's verify the indices.
For $k=3$: $S_3 - e_1 S_2 + e_2 S_1 - 3e_3 = 0$.
Wait, the coefficient of $e_3$ is $3$ only in the specific Newton sum formula for $S_3$.
The general recurrence relation for roots of a polynomial $t^n - e_1 t^{n-1} + \dots + (-1)^n e_n = 0$ is:
$S_k - e_1 S_{k-1} + e_2 S_{k-2} - \dots + (-1)^{n-1} e_{n-1} S_{k-n+1} + (-1)^n n e_n = 0$ for $k=n$.
And for $k > n$:
$S_k - e_1 S_{k-1} + e_2 S_{k-2} - \dots + (-1)^n e_n S_{k-n} = 0$.

Here $n=3$.
So for $k=4$:
$S_4 - e_1 S_3 + e_2 S_2 - e_3 S_1 = 0$.
Let's check the signs.
The polynomial is $t^3 - e_1 t^2 + e_2 t - e_3 = 0$.
So $x^3 = e_1 x^2 - e_2 x + e_3$.
Summing over $x, y, z$:
$S_3 = e_1 S_2 - e_2 S_1 + 3e_3$. This matches the $k=3$ formula.
Multiply by $x$: $x^4 = e_1 x^3 - e_2 x^2 + e_3 x$.
Summing over $x, y, z$:
$S_4 = e_1 S_3 - e_2 S_2 + e_3 S_1$.
This matches the recurrence $S_4 - e_1 S_3 + e_2 S_2 - e_3 S_1 = 0$.

Let's plug in the values we found.
$e_1 = 6$
$e_2 = 11$
$e_3 = 6$
$S_1 = 6$
$S_2 = 14$
$S_3 = 36$

$S_4 = 6(36) - 11(14) + 6(6)$
$S_4 = 216 - 154 + 36$
$216 + 36 = 252$
$252 - 154 = 98$.

Let's double check the calculation.
$6 \times 36 = 216$.
$11 \times 14 = 154$.
$6 \times 6 = 36$.
$216 - 154 = 62$.
$62 + 36 = 98$.

Wait, $216 - 154$:
$216 - 100 = 116$
$116 - 50 = 66$
$66 - 4 = 62$.
$62 + 36 = 98$.
Yes.

Let's re-verify the roots calculation.
$P(t) = t^3 - 6t^2 + 11t - 6$.
Roots are $1, 2, 3$.
$1+2+3 = 6$.
$1\cdot 2 + 2\cdot 3 + 3\cdot 1 = 2 + 6 + 3 = 11$.
$1\cdot 2\cdot 3 = 6$.
Matches $e_1, e_2, e_3$.
$S_4 = 1^4 + 2^4 + 3^4 = 1 + 16 + 81 = 98$.

Is there any ambiguity about "Real numbers"?
The roots $1, 2, 3$ are real.
If the roots were complex, $S_4$ would still be determined by the recurrence, but the problem states $x, y, z$ a