If $$x^4+x^3+x^2+x+1=0$$ then what's the value of $x^5$ ??

I thought it would be $-1$ but it does not satisfy the equation


3 Answers 3


Well, we have $x^5-1=(x-1)(x^4+x^3+x^2+x+1)=(x-1)0=0$ so that $x^5=1$


$x^4+x^3+x^2+x+1 = 0$

Multiply everything by $x$

$x^5+x^4+x^3+x^2+x = 0$

Add $1$ to both sides:

$x^5+x^4+x^3+x^2+x+1 = 1$

$x^5+(x^4+x^3+x^2+x+1) = 1$

$x^5+(0) = 1$



The left side of the equation is $P= (x^4 +x^2) + (x^3 + x) +1 = 2$

In the equation $x^2(x^2+1) +1 > 0$, there is no such real number for $x$. Logically, a false statement implies anything, so $x^5$ = anything you wish .

M Andrews' correct argument above can be modified to see that $x^5-1 =(x-1)(x^4+x^3+x^2+x+1)$, so the roots of $P=0$ are the 4 other complex 5th roots of unity (which are not x=1) namely $x=e^{2m\pi i}/5$, so $m=1,2,3,4$, and so on.

For instance, for $m=1$ $x= \cos(2\pi/5) + i \sin(2\pi/5)$ . So $x^5 =1 $, provided you allow complex numbers as roots of P.

  • $\begingroup$ Is pie $\pi$? I need to edit $\endgroup$
    – Xetrov
    May 12, 2017 at 7:12

Not the answer you're looking for? Browse other questions tagged or ask your own question.