3
$\begingroup$

I am reading a Wikipedia article on Ramanujan's sum. The article states that:

It follows from the identity $x^q − 1 = (x − 1)(x^{q−1} + x^{q−2} + ... + x + 1)$ that the sum of the nth powers of All the qth roots of unity is \begin{cases}0&\;{\mbox{ if }}q\nmid n\\q&\;{\mbox{ if }}q\mid n\\\end{cases}

I think I understand the case for $q|n$. Every $q$th root of unity raised to a multiple of $q$ would equal $1$ so the sum is $1+1+1+...+1$ ($q$ summands of $1$). Right?

I also understand that the sum of all the $q$th roots of unity (not raised to any power) is $0$.

I have an intuitive hazy notion that if $q$ does not divide $n$ then the roots upon being raised to the $n$th power are mapped bijectively onto themselves so the sum is $0$.

How can I prove this? How does this follow from the given identity?

$\endgroup$
3
$\begingroup$

Your understanding for the case $q|n$ is right.

I use $x$ to denote a $q$-th root of unity for the rest of this answer. Note that $x^n = 1 \iff q | n$. If $q\nmid n$, $x^n \ne 1$. Substitute $x^n$ into the given identity and divide both sides by $x^n-1$.

$$\sum_{k=0}^{q-1} x^{nk} = \frac{x^{qn}-1}{x^n-1} = \frac{(x^{q})^n-1}{x^n-1} = \frac{1-1}{x^n-1}=0.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.