How can I apply Newton's sums to solve this problem? 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?

Here is another useful application of the AM-GM inequality(didn't think of it, did you)
Tricky factorization:
$$0=-(a^7+b^7+c^7+d^7)$$$$=7(b+c)(c+d)(d+a)E$$ where $E=(b^2+c^2+d^2+bc+cd+da)^2+bcd(b+c+d)$. However, $$4E=((b+c)^2+(c+d)^2+(b+d)^2)^2)-4abcd$$$$=((b+a)^2+(c+a)^2+(a+d)^2)^2)-4abcd$$$$=(3a^2+2a(b+c+d)+b^2+c^2+d^2)^2-4abcd$$ The last expression reduces to $$(a^2+b^2+c^2+d^2)^2-4abcd$$ However we see that $$(a^2+b^2+c^2+d^2)^2-4abcd \geq 12|abcd| \geq 0$$ from the AM-GM inequality. We conclude that $4E \geq 0$ with equality if and only if $|a|=|b|=|c|=|d|=0$. Otherwise one of $b+c,b+d,c+d$ must be $0$. Hence, the result follows.
