Number of values of Z(real or complex) satisying the given system of equations..? 
Number of values of Z(real or complex) simultaneously satisfying the
  system of equations 
${1+Z^2+Z^3+....+Z^{17}}$=$0$
and 
${1+Z^2+Z^3+....+Z^{13}}$=$0$ is...?

My attempt : $Z^{18}-1$$=$$ (Z-1)$ ${(1+Z^2+Z^3+....+Z^{17})}$ (Algebraic identity) 
Also, ${1+Z^2+Z^3+....+Z^{17}}$=$0$ 
Therefore, $Z^{18}=1$ -----------(i)
Similarly, $Z^{14}=1$ -----------(ii) 
On dividing the above two equations, I get $Z^4$=$1$ which has four solutions 1,-1,i and -i; therefore, the number of required values of Z must be 4. 
On plugging these values back in the equation, it can be clearly seen that 1,i and -i don't satisfy the equation only -1 does. Why did I arrive mathematically at these values only to have them rejected?
 A: When you manipulate equations in non-reversible ways (like multiplying by $z-1$, or dividing) you are potentially introducing additional roots. In this case the roots of the original system are among the roots of $z^4=1$, but not all roots of $z^4=1$ necessarily satisfy the original system.
Alternative hint to solve it: $\;\gcd(z^m-1,z^n-1)=z^{\gcd(m,n)}-1\,$ and of course $\gcd(18,14)=2\,$.

[ EDIT ]  Combining the same $\gcd$ idea with OP's approach:


*

*$z^{18}=1 \implies \left(z^{18}\right)^3=z^{54}=1$

*$z^{14}=1 \implies \left(z^{14}\right)^4=z^{56}=1$
Dividing the two equations gives $z^2=1\,$, and dropping the extraneous root $z=1$ which was introduced by the multiplication with $z-1$ leaves $z=-1\,$.
A: Another approach might be to subtract one equation from the other to find 
$$
\begin{align}
Z^{14}+Z^{15}+Z^{16}+Z^{17}&=0 \\  
Z^{14}(1+Z+Z^2+Z^3)&=0 \\
Z^{14}(Z+1)(Z-i)(Z+i)&=0 \\
\end{align}
$$
We still end up with three extraneous solutions!  It really goes to the point made by Stahl as a comment.  At the very first step all we know is that a solution to our problem will satisfy $Z^{14}+Z^{15}+Z^{16}+Z^{17}=0$ as well, but we don't know that every solution to $Z^{14}+Z^{15}+Z^{16}+Z^{17}=0$ satisfies our original problem.
