$a+b+c+d+e$ divides $a^5+b^5+c^5+d^5+e^5-5abcde$ 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 $a^5+b^5+c^5+d^5+e^5-5abcde$.
I'm reminded of the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. But for $5$th degree, how can I find a factorization for $a^5+b^5+c^5+d^5+e^5-5abcde$?
 A: Hint: firts note that; if $a(b+c)+b(c+d)+c(d+e)+d(e+a)+e(a+b)=0$, then  $$(a+b+c+d+e)^2=a^2+b^2+c^2+d^2+e^2$$
Now take $P(x)=x^5+kx^4+rx^3+sx^2+tx+u$, with roots $a,b,c,d,e$ then  from Viète’s
Relations; 
$\boxed{ u=-abcde}$, 
$\boxed{k=-(a+b+c+d+e)}$,
$\boxed{r=ae+be+ce+de+ab+ac+ ad+bc+bd+cd=0}$ 
Take $m=a^4+b^4+c^4+d^4+e^4$ and How $P(a)+P(b)+P(c)+P(d)+P(e)=0$, then:
$${a^5+b^5+c^5+d^5+k(m)+s(a^2+b^2+c^2+d^2+e^2)+t(a+b+c+d+e)+5u=0}$$
then 
$${a^5+b^5+c^5+d^5-5u=-k(m)-s(a^2+b^2+c^2+d^2+e^2)-t(a+b+c+d+e)}$$
$$\implies{a^5+b^5+c^5+d^5-5abcde=(a+b+c+d+e)\cdot M}$$
$$\implies(a+b+c+d+e)|(a^5+b^5+c^5+d^5+e^5-5abcde)$$
A: If you use the following notations:
$$s_1=a+b+c+d+e$$
$$s_2=ab+ac+ad+ae+bc+bd+be+cd+ce+de$$
$$s_3=abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde$$
$$s_4=abcd+abce+abde+acde+bcde$$
$$s_5=abcde$$
Then we have:
$$a^5+b^5+c^5+d^5+e^5=\left(\left(s_1\left(s_1^2-2s_2\right)-s_1s_2+3s_3\right)s_1-s_2\left(s_1^2-2s_2\right)+s_3s_1-4s_4\right)s_1-s_2\left(s_1\left(s_1^2-2s_2\right)-s_1s_2+3s_3\right)+s_3\left(s_1^2-2s_2\right)-s_4s_1+5s_5$$
The given condition is in this notation:
$$s_2=0$$
From that it follows that:
$$a^5+b^5+c^5+d^5+e^5-5abcde=s_1\left(s_1^4+5s_1s_3-5s_4\right)$$
And the desired result follows.
