Let $n$ be a positive integer and a complex number with unit modulus is a solution of the equation $z^n+z+1=0$. Prove that $n $ can't be $196$.

Question

Let $n$ be a positive integer and a complex number with unit modulus is a solution of the equation $z^n+z+1=0$. Prove that $n $ can't be $196$.

The above question has been bothering me since a long time. I 've tried using the Euler's form for $z $ and have obtained $\sin 2nx=-0.5$. I don't know how to use that. Would someone help me to solve this problem?

Thanks in advance.

Answer 1

Let $$z^{196}+z+1=0$$

Then we have $$z^{196}= -z-1$$

Thus $$|z|^{196} =|-z-1|$$

Since $|z|=1$ we get $|z+1|=1$

Let $z=x+iy$ then we have $z+1=(x+1)+iy$ so $|z+1|^2 =(x+1)^2+y^2 =1$

That is $x^2+y^2+2x+1=1$ which implies $x=-1/2$

since we have $|z|=1$ we have $z=-1/2 \pm i {\sqrt 3}/2 =e^{\pm 2\pi i/3}$

which does not satisfy $$z^{196}+z+1=0$$

Answer 2

$z^n+z+1=0$ implies $1=|z|^n=|z^n|=|z+1|$.

If moreover $|z|=1$, then $z$ is a primitive cubic root of $1$ and so $z^2+z+1=0$.

(Indeed, $|z+1|=1$ and $|z|=1$ define two circles which intersect at the primitive cubic roots of $1$.)

Therefore, $z^n=z^2$ and so $z^{n-2}=1$. Thus, $n \equiv 2 \bmod 3$. However, $196 \equiv 1 \bmod 3$.