Show that either $1+\alpha+\alpha^2+.....+\alpha^{p-1}=0$ or $1+\alpha+\alpha^2+.....+\alpha^{q-1}=0$,but not both together. Let a complex number $\alpha,\alpha\neq1$,be a root of the equation $z^{p+q}-z^p-z^q+1=0$,where $p$ and $q$ are distinct primes.Show that either $1+\alpha+\alpha^2+.....+\alpha^{p-1}=0$ or $1+\alpha+\alpha^2+.....+\alpha^{q-1}=0$,but not both together.
Since $\alpha$ is a root of $z^{p+q}-z^p-z^q+1=0$.Therefore,$\alpha^{p+q}-\alpha^p-\alpha^q+1=0$ factorised to $(\alpha^p-1)(\alpha^q-1)=0$,but since $\alpha\neq1$
Therefore,either $1+\alpha+\alpha^2+.....+\alpha^{p-1}=0$ or $1+\alpha+\alpha^2+.....+\alpha^{q-1}=0$ or both equal to zero.Because we have studied that if $ab=0$,then either $a=0$ or $b=0$ or both equal to zero.But this is in contradiction to what is to be proved.My query is why cannot both be zero? 

If $a$ is a root of $1+z+z^2+\cdots+z^{m-1}=0\  \ \ \  (1),a\ne1$ 
and is a root of $\dfrac{z^m-1}{z-1}=0$
If $a$ is a root of $1+z+z^2+\cdots+z^{n-1}=0\  \ \ \  (2),a$ is a root of is a root of $\dfrac{z^n-1}{z-1}=0$
Now Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ says $(a^m-1,a^n-1)=a^{(m,n)}-1$
$$\implies\left(\dfrac{a^m-1}{a-1},\dfrac{a^n-1}{a-1}\right)=\dfrac{a^{(m,n)}-1}{a-1}$$
If $(m,n)=1,$
$$\left(\dfrac{a^m-1}{a-1},\dfrac{a^n-1}{a-1}\right)=1$$
$\implies(1),(2)$ can not have any root in common
