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.

  • $\begingroup$ I am a little confused about what you have under that radical, I came up with -3 instead of 5 $\endgroup$ – imranfat May 7 '14 at 16:26
  • $\begingroup$ It will be $5$. $\endgroup$ – Samrat Mukhopadhyay May 7 '14 at 16:27
  • $\begingroup$ @imranfat: It can be rewritten as $$\left(a+\frac1a\right)^2+a+\frac1a-1=0.$$ $\endgroup$ – Cameron Buie May 7 '14 at 16:33
  • $\begingroup$ The first equation will have $4$ solutions, isn't the case that the second equation also will have $4$ solutions? If this is right, there is no sense asking "the value" of the second equation, there is $4$ possible values. $\endgroup$ – Integral May 7 '14 at 16:36
  • $\begingroup$ @Integral No, the second equation has $4m$ solutions, not $4$. $\endgroup$ – user26486 May 7 '14 at 16:38

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 $m\not\equiv 0\pmod 5$


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.

  • $\begingroup$ Just to show that the answer can be found without having to find the roots of the original equation. $\endgroup$ – Mark Bennet May 7 '14 at 16:44

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 $$\frac{a^{5m}-1}{a^{2m}(a-1)}-1=\left\{\begin{array}{rl} 4 & m=0 \mod 5\\ -1 & \mbox{else} \end{array} \right.$$


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