Let $a$, $b$, $c$, $d$ be the roots of $x^4 + x + 1 = 0$. Find $a^4 + b^4 + c^4 + d^4$.

Question

Let $p(x) = x^4 + x + 1 = 0$, and let $a$, $b$, $c$, $d$ be its roots. Find $a^4 + b^4 + c^4 + d^4$.

I have no idea how to start solving this problem.

Answer 1

For any of the roots, $r^4=-r-1$ so that the requested sum is $-a-b-c-d-4$, and by the Vieta's formula, the sum of the roots is $0$.

Answer 2

As an alternative, let consider

$$(x-a)(x-b)(x-c)(x-d)=$$ $$=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2\\-(abc+abd+acd+bcd)x+abcd$$

then

• $S_1=a+b+c+d=0$
• $S_2=ab+ac+ad+bc+bd+cd=0$
• $S_3=abc+abd+acd+bcd=-1$
• $S_4=abcd=1$

and by Newton's sums we have that

• $P_1=a+b+c+d=S_1=0$
• $P_2=a^2+b^2+c^2+d^2=S_1P_1-2S_2=0$
• $P_3=a^3+b^3+c^3+d^3=S_1P_2-S_3P_1+3S_3=-3$
• $P_4=a^4+b^4+c^4+d^4=S_1P_3-S_2P_2+S_3P_1-4S_4=-4$