Given x_1,x_2,x_3,x_4 the roots of p(x) = x^4+x^3+x^2+x+1 = 0 calculate x_1^8+x_2^(18)+x_3^(28)+x_4^(38) ?

2 Answers
Jun 12, 2017

x_1^8+x_2^18+x_3^28+x_4^38 = -1

Explanation:

Given:

p(x) = x^4+x^3+x^2+x+1

Note that:

(x-1)p(x) = (x-1)(x^4+x^3+x^2+x+1) = x^5-1

So we can tell that the zeros of p(x) are all 5th roots of 1.

So if the zeros of p(x) are x_1, x_2, x_3 and x_4 then:

x_1^5=x_2^5=x_3^5=x_4^5=1

So:

x_1^8+x_2^18+x_3^28+x_4^38 = x_1^3+x_2^3+x_3^3+x_4^3

Now we have a symmetric polynomial in the zeros, we can express it in terms of the elementary symmetric polynomials that form the coefficients of p(x)...

Note:

p(x) = (x-x_1)(x-x_2)(x-x_3)(x-x_4)

color(white)(p(x)) = x^4-(x_1+x_2+x_3+x_4)x^3+(x_1x_2+x_2x_3+x_3x_4+x_4x_1)x^2-(x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2)x+x_1x_2x_3x_4

Equating coefficients, we have:

{ (x_1+x_2+x_3+x_4 = -1), (x_1x_2+x_2x_3+x_3x_4+x_4x_1=1), (x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2=-1), (x_1x_2x_3x_4=1) :}

Now:

x_1^2+x_2^2+x_3^2+x_4^2

=(x_1+x_2+x_3+x_4)^2-2(x_1x_2+x_2x_3+x_3x+x_4x_1+x_1x_3+x_2x_4)

=(-1)^2-2(1)

=-1

Note that:

-1 = (x_1+x_2+x_3+x_4)(x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1x_3+x_2x_4)

color(white)(-1) = (x_1x_2^2+x_2x_3^2+x_3x_4^2+x_4x_1^2+x_1x_3^2+x_2x_4^2+x_1^2x_2+x_2^2x_3+x_3^2x_4+x_4^2x_1+x_1^2x_3+x_2^2x_4)+3(x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2)

color(white)(-1) = (x_1x_2^2+x_2x_3^2+x_3x_4^2+x_4x_1^2+x_1x_3^2+x_2x_4^2+x_1^2x_2+x_2^2x_3+x_3^2x_4+x_4^2x_1+x_1^2x_3+x_2^2x_4)+3(-1)

Hence:

x_1x_2^2+x_2x_3^2+x_3x_4^2+x_4x_1^2+x_1x_3^2+x_2x_4^2+x_1^2x_2+x_2^2x_3+x_3^2x_4+x_4^2x_1+x_1^2x_3+x_2^2x_4 = 2

Then:

1 = (x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2)

color(white)(1)=(x_1^3+x_2^3+x_3^3+x_4^3)+(x_1x_2^2+x_2x_3^2+x_3x_4^2+x_4x_1^2+x_1x_3^2+x_2x_4^2+x_1^2x_2+x_2^2x_3+x_3^2x_4+x_4^2x_1+x_1^2x_3+x_2^2x_4)

color(white)(1)=(x_1^3+x_2^3+x_3^3+x_4^3)+2

So:

x_1^3+x_2^3+x_3^3+x_4^3 = -1

Jun 12, 2017

See below.

Explanation:

P(X)= (X^5-1)/(X-1) so the roots of

P(X)=0 are x_k=e^(i (2kpi)/5) for k=1,2,3,4

but

8 equiv 3 mod 5
18 equiv 3 mod 5
28 equiv 3 mod 5
38 equiv 3 mod 5

so

x_1^8+x_2^(18)+x_3^(28)+x_4^(38) = x_1^3+x_2^3+x_3^3+x_4^3=-1

because

x_0+x_1+x_2+x_3+x_4 = 0
x_0^2+x_1^2+x_2^2+x_3^2+x_4^2 = 0
x_0^3+x_1^3+x_2^3+x_3^3+x_4^3 = 0
x_0^4+x_1^4+x_2^4+x_3^4+x_4^4 = 0

Don't forget that x_0 = 1