Find the value of $a^{2m}+a^m+{1\over a^m}+{1\over a^{2m}}$ Let $a$ be a complex number such that $a^2+a+{1\over a}+{1\over a^2}+1=0$
Let $m$ be a positive integer, find the value of $a^{2m}+a^m+{1\over a^m}+{1\over a^{2m}}$
My approach: I factorized equation 1 which yielded $(a+{1\over a})={-1^+_-(5^{1\over2})\over2}$
I don't know if it is of any help.
 A: Hint: Multiplying the first equation by $a^2(a-1)$, we get $a^5=1$ and so $a=e^{\frac{2ki\pi}5}$ for $k=1,2,3,4$
So second expression is $2\cos\frac{2kmi\pi}5+2\cos\frac{4kmi\pi}5$ and so is :
$4$ if $m\equiv 0\pmod 5$
$-1$ if $m\not\equiv 0\pmod 5$
A: Multiply through by $a^2$ to obtain $$a^4+a^3+a^2+a+1=0$$Multiply by $(a-1)$ to get $$a^5-1=0; a^5=1$$ since $a\neq 1$
Now we can consider the exponents mod $5$.
If $m$ is a multiple of $5$ we have one case. If it is not a multiple of $5$ we have another.
In the first case all the terms are equal to $1$, giving the answer $4$.
In the second case the exponents $m, 2m, -m, -2m$ are all distinct and non-zero mod $5$ [if, say $m\equiv -2m$, we'd have $5|3m$, whence $5|m$ contradiction] and are therefore equivalent in some order to $1,2,3,4$
Then $a^4+a^3+a^2+a=-1$ from the first equation of this answer.
A: The equation can be transformed to $$a^4+a^3+a^2+a+1=0\Rightarrow a=\exp(2\pi k/5),\ k=1,2,3,4$$ Hence, the expression becomes $$\frac{a^{5m}-1}{a^{2m}(a-1)}-1=\left\{\begin{array}{rl}
4 & m=0 \mod 5\\
-1 & \mbox{else}
\end{array}
\right.$$
