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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?

Arishta
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    What you have shown is that if $Z$ satisfies your two original equations (and $Z\neq 0$, you have to assume this to do your division step), then $Z^4 = 1$. So the only possibilities for $Z$ are $\pm 1,\pm i$, but you are not guaranteed that these are solutions; see dxiv's answer. – Stahl Jun 14 '17 at 16:23

2 Answers2

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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 $z=-1\,$.

dxiv
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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.

sharding4
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