Given $x_1,x_2,x_3,x_4$ real numbers such that $x_1+x_2+x_3+x_4 = 0$ and $x_1^7+x_2^7+x_3^7+x_4^7 = 0,$ how can I use symmetric functions and Newton's sums to prove that $x_1(x_1+x_2)(x_1+x_3)(x_1+x_4)=0$?
This is what I have so far:
Note that $x_1 = -(x_2+x_3+x_4)$ and the relation between Newton's sums, elementary symmetric polynomials, and power sums gives us that $P_7-S_1 P_6+S_2 P_5-S_3 P_4+S_4P_3=0,$ where $P_k$ denotes the $kth$ power sum and $S_k$ denotes the $kth$ elementary symmetric polynomial of $x_1,x_2,x_3,x_4.$
We want to show that $(x_2+x_3+x_4)(x_2+x_3)(x_2+x_4)(x_3+x_4) = 0.$
But how?
Best Answer
The key to this problem is to use all the Newton's Sums relations, not just the one with $S_7$.
Let $(x-x_1)(x-x_2)(x-x_3)(x-x_4) = x^4+a_1x^3+a_2x^2+a_3x+a_4$ be a $4$th degree polynomial with roots $x_1,x_2,x_3,x_4$, and let $S_n = x_1^n+x_2^n+x_3^n+x_4^n$ be the sum of the $n$-th powers of the roots of the polynomial.
We are given that $S_1 = S_7 = 0$. Then, Newton's Sums gives us the following:
$S_1+a_1 = 0 \leadsto a_1 = 0$
$S_2+a_1S_1+2a_2 = 0 \leadsto S_2 = -2a_2$
$S_3+a_1S_2+a_2S_1+3a_3 = 0 \leadsto S_3 = -3a_3$
$S_4+a_1S_3+a_2S_2+a_3S_1+4a_4 = 0 \leadsto S_4 = 2a_2^2-4a_4$
$S_5+a_1S_4+a_2S_3+a_3S_2+4S_1 = 0 \leadsto S_5 = 5a_2a_3$
Let's skip the relation for $S_6$ since it isn't needed.
$S_7+a_1S_6+a_2S_5+a_3S_4+4S_3 = 0 \leadsto 7(a_2^2-a_4)a_3 = 0$
So, either $a_3 = 0$ or $a_2^2 = a_4$.
If $a_2^2 = a_4$, then $S_4 = 2a_2^2-4a_4 = -2a_2^2 \le 0$. But since $x_1,x_2,x_3,x_4$ are real, we must have $S_4 = x_1^4+x_2^4+x_3^4+x_4^4 \ge 0$. Thus, $S_4 = 0$, which gives us $x_1 = x_2 = x_3 = x_4 = 0$, and thus, $x_1(x_1+x_2)(x_1+x_3)(x_1+x_4)=0$.
If $a_3 = 0$, then $x_1,x_2,x_3,x_4$ are roots of the polynomial $x^4+a_2x^2+a_4$. Since this polynomial is even, either $x_1 = 0$ or $x_1 = -x_i$ for some $i = 2,3,4$, and thus, $x_1(x_1+x_2)(x_1+x_3)(x_1+x_4)=0$.
Therefore, $x_1(x_1+x_2)(x_1+x_3)(x_1+x_4)=0$, as desired.