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?

  • $\begingroup$ Try defining the polynomial (x-a)(x-b)(x-c)(x-d) and plugging in x=-a? $\endgroup$ Jun 17, 2015 at 0:22
  • $\begingroup$ @FarazMasroor I don't really see how such a polynomial can use the condition with $7$-th power. $\endgroup$ Jun 17, 2015 at 0:24
  • 2
    $\begingroup$ Neither do I to be completely honest. Naive me would write out all those painful newtons sums and hope for the best. $\endgroup$ Jun 17, 2015 at 0:27
  • 1
    $\begingroup$ You can simplify the problem to proving that if $(1+b+c)^7 = 1+b^7+c^7$, then $(1+b+c+bc)(b+c)=0$ by eliminating $d$ from both equations and taking $a\ne 0$ (because $a=0$ is not interesting) and rescaling to $a=1$. After that, I don't see an elegant way of continuing. $\endgroup$ Jun 17, 2015 at 0:41

2 Answers 2


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.


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.


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