Question

Let a complex number $\alpha ,\alpha \ne 1$, be a root of the equation $z^{p+q}-z^p-z^q+1=0$, where p, q are distinct primes, then either $1+\alpha +{\alpha }^2+..+{\alpha }^{p-1}=0$ or $1+\alpha +{\alpha }^2+...+{\alpha }^{q-1}=0$, but not both together. If this is true enter 1, else enter 0.

Answer 1

$z^{p+q}-z^p-z^q+1=0$

$\Rightarrow z^p(z^q-1)-1(z^q-1)=0$
$\Rightarrow (z^p-1)(z^q-1)=0$

$\Rightarrow z^p-1=0$ or $z^q-1=0$
$z^p=1$ gives $pth$ roots of unity.
Now we know that sum of $nth$ roots of unity is zero.
$\therefore 1+\alpha +{\alpha }^2...{\alpha }^{p-1}=0$
Similarly
$z^q-1=0$
$z^q=1$ gives $qth$ roots of unity.
Now we know that sum of $nth$ roots of unity is zero.
$\therefore 1+\alpha +{\alpha }^2...{\alpha }^{q-1}=0$

Hence, either $1+\alpha +{\alpha }^2...{\alpha }^{p-1}=0$ or $1+\alpha +{\alpha }^2...{\alpha }^{q-1}=0$