Suppose $γ$ is a $k$th root of unity that satisfies a quadratic equation $z^2−mz−n=0$ with $m,n\in\mathbb{Z}$. Then $k=3,4$ or $6$. 
Let $k\in\mathbb{Z}$ with $k>2$ and suppose $\gamma$ is a $k$th root of unity that satisfies a quadratic equation $z^2-mz-n=0$ with $m,n\in\mathbb{Z}$. Then $k=3,4$ or $6$.  

My knowledge on algebra and number theory is poor, I just need this lemma to complete the computation of the analytic automorphism groups of complex tori.
Can you help me prove this lemma with the knowledge of field theory and Euler phi function as small as possible? Actually I forget almost everything on field theory and have never learned number theory. Thanks in advance!   
 A: The problem statement is somewhat sloppy as we shall see:
If $\gamma$ is a $k$th root of unity, $k\ge 1$, and root of a quadratic polynomial $X^2+aX+b\in\mathbb Z[X]$ then $X^2+aX+b$ divides $X^k-1$ or vice versa. As $k\ge 2$, we may restrict to the first case. It follows that $b\mid (-1)$, i.e. $b=\pm1$. Then since $|\gamma^2\pm1|\le 2$ and $|a\gamma|=|a|$, it follows that $a\in\{-2,-1,0,1,2\}$. Then $\Re\gamma = -\frac a2\in\{-1,-\frac12,0,\frac12,1\}$, which implies that $\gamma$ is a primitive second or third or fourth or sixth or first root of unity. 
For each of these cases we can indeed name a quadratic polynomial as required:


*

*$\gamma$ is a primitive first root of unity, i.e. $\gamma=1$: Then $\gamma^2-1=0$.

*$\gamma$ is a primitive second root of unity, i.e. $\gamma=-1$: Then $\gamma^2-1=0$.

*$\gamma$ is a primitive third root of unity, i.e. $\gamma=\frac{-1\pm i\sqrt 3}2$: Then $\gamma^2+\gamma+1=0$.

*$\gamma$ is a primitive fourth root of unity, i.e. $\gamma=\pm i$: Then $\gamma^2+1=0$.

*$\gamma$ is a primitive sixth root of unity, i.e. $\gamma=\frac{1\pm i\sqrt 3}2$: Then $\gamma^2-\gamma+1=0$.


Of course, a number like $\gamma=1$ is also a $k$th root of unity for $k=42$ or $k=666$ (though not a primitive such root - that word is missing from the problem statement).  
A: Let $\displaystyle u=e^{i\psi}=\cos\psi+i\sin\psi$ be a $k$ th root of unity $\implies (e^{i\psi})^k=1=e^{i\psi k}=1$
So, we have $\displaystyle (\cos\psi+i\sin\psi)^2-m(\cos\psi+i\sin\psi)+n=0$
$\displaystyle\implies \cos2\psi+i\sin2\psi-m(\cos\psi+i\sin\psi)+n=0\  \ \ \ (1)$
Equating the imaginary parts,  $\displaystyle\sin2\psi-m\sin\psi=0\iff 
\sin\psi(2\cos\psi-m)=0$
If $\displaystyle\sin\psi=0,\cos\psi=\pm1, u=\pm1\implies k=2$ but given that $k>2$
$(1)$ becomes $$1-m(\pm1)+n=0\iff\mp m=n+1$$
Else $\displaystyle\cos\psi=\frac m2\implies -1\le\frac m2\le1\iff -2\le m\le2$
As $m\in\mathbb{Z}, m=0,\pm1,\pm2\  \ \ \ (1)$
Now check for values of $m$ each of which gives exactly one value of $\sin\psi$ 
A: If you're looking for primitive roots of unity, you're looking for those n for which the cyclotomic polynomial has degree 2, i.e. $\phi(n)=2$.
