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
George C.
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 #5#th 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#

Cesareo R.
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#

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