If z is a complex number, then the number of common roots of the equations z^1985+z^100+1=…

Question

If $z$ is a complex number, then the number of common roots of the equations $z^{1985}+z^{100}+1=0$ and $z^3+2z^2+2z+1=0$, is equal to:

Accepted Answer

Factor the cubic: $z^3+2z^2+2z+1=(z+1)(z^2+z+1)$, whose roots are $z=-1,\ \omega,\ \omega^2$.

Check $z=-1$: $(-1)^{1985}+(-1)^{100}+1=-1+1+1=1
eq0$. Not a root of the first equation.

Check $z=\omega$: $1985\equiv2\pmod3$ and $100\equiv1\pmod3$, so $\omega^{1985}+\omega^{100}+1=\omega^2+\omega+1=0$.

Check $z=\omega^2$: $(\omega^2)^{1985}=\omega^{3970}$, and $3970\equiv1\pmod3$, so $=\omega$. $(\omega^2)^{100}=\omega^{200}$, and $200\equiv2\pmod3$, so $=\omega^2$. Sum: $\omega+\omega^2+1=0$.

So $\omega$ and $\omega^2$ are both common roots: 2 common roots.

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