Mathematical Reflections 2013, Issue 2 - Problem S259

Problem:
Let $a,b,c,d,e$ be integers such that $$a(b+c)+b(c+d)+c(d+e)+d(e+a)+e(a+b)=0.$$ Prove that $a+b+c+d+e$ divides
$a^5 + b^5 + c^5 + d^5 + e^5 - 5abcde$.

Proposed by Titu Andreescu.

Solution:
Suppose that $a,b,c,d,e$ are the five roots $\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5$ of a fifth degree polynomial $P(x)$.
Let $$\sigma_k=\sum_{i=1}^5 \alpha^k_i, \qquad s_k=\sum_{1 \leq j_1 < j_2 < \ldots < j_k \leq 5} \alpha_{j_1}\alpha_{j_2}\cdots\alpha_{j_k}.$$ With this notation, we know that $s_2=0$ and we want to prove that $\sigma_1$ divides $\sigma_5-5s_5$. We have $$P(x)=x^5-s_1x^4+s_2x^3-s_3x^2+s_4x-s_5,$$ so
$$\begin{array}{lcl} P(\alpha_1)&=&\alpha^5_1-s_1\alpha^4_1+s_2\alpha^3_1-s_3\alpha^2_1+s_4\alpha_1-s_5=0 \\ P(\alpha_2)&=&\alpha^5_2-s_1\alpha^4_2+s_2\alpha^3_2-s_3\alpha^2_2+s_4\alpha_2-s_5=0 \\ P(\alpha_3)&=&\alpha^5_3-s_1\alpha^4_3+s_2\alpha^3_3-s_3\alpha^2_3+s_4\alpha_3-s_5=0 \\
P(\alpha_4)&=&\alpha^5_4-s_1\alpha^4_4+s_2\alpha^3_4-s_3\alpha^2_4+s_4\alpha_4-s_5=0 \\
P(\alpha_5)&=&\alpha^5_5-s_1\alpha^4_5+s_2\alpha^3_5-s_3\alpha^2_5+s_4\alpha_5-s_5=0. \end{array}$$
Summing up the columns, we get $$\sigma_5-s_1\sigma_4+s_2\sigma_3-s_3\sigma_2+s_4\sigma_1-5s_5=0.$$ Since $s_1=\sigma_1, s_2=0$ and $\sigma_2=\sigma^2_1-2s_2=\sigma^2_1$ we obtatin $$\sigma_5-5s_5=\sigma_1(\sigma_4+s_3\sigma_1-s_4),$$ hence $\sigma_1|(\sigma_5-5s_5)$.