# $a+b+c+d+e$ 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 $a^5+b^5+c^5+d^5+e^5-5abcde$?

• $a=-b-c-d-e$ doesn't yield $0$, so it won't be that easy. Nov 18, 2014 at 4:05
• I'm having difficulty with this one as well, perhaps we need to "add zero" to make it factorable, which in this case would include multiples of the first line as well. If this was how the problem was worded it is highly likely to be untrue in general without the stated condition. Nov 18, 2014 at 4:12

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 $${a^5+b^5+c^5+d^5-5u=-k(m)-s(a^2+b^2+c^2+d^2+e^2)-t(a+b+c+d+e)}$$ $$\implies{a^5+b^5+c^5+d^5-5abcde=(a+b+c+d+e)\cdot M}$$

$$\implies(a+b+c+d+e)|(a^5+b^5+c^5+d^5+e^5-5abcde)$$

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.