Number of elements which are cubes/higher powers in a finite field.

This question is a slight generalization of This Question.

How many elements are there in a finite field of order $q$ which are :

• Squares.
• Cubes.
• Higher powers.

I mean :

How many elements are there which are squares...

How many elements are there which are cubes...

how many elements are there which are higher powers...

What all I could see is :

For finding elements which are cubes I would consider :

$\eta : F^*\rightarrow F^*$ sending $x\rightarrow x^3$

For this If I know the kernel of $\eta$ then I would have $F^*/Ker(\eta) \cong \{x^3 : x\in F^*\}$

Which says that :

No of elements which are cubes are $\dfrac{|q-1|}{|Ker (\eta)|}$.

Now, Kernel of $\eta$ would be $\{x\in F^* : x^3=1\}$

All I know is if there is some subgroup (I am mentioning $\{1,x,x^2\}$ for $\{x\in F^* : x^3=1\}$)

Then It would be unique as a finite cyclic group can not have two subgroups of same order.

So, Now the problem is how to see for existence as we are through with uniqueness.

Suppose I prove uniqueness then I would say :

Number of elements of $F^*$ which are squares are :

• $\dfrac{q-1}{3}$ If $F^*$ have an element of order $3$
• $q-1$ If $F^*$ have no element of order $3$

(I believe this would be totally dependent of nature of $q$)

Now the question is how do i make sure of existence with given nature of $q$.

I would be Thankful If some one can help me to see this and I would be happy to see further generalization.

How many elements of order $n$ are there in a Finite field of cardinality $q$

Thank you :)

This problem becomes easy if you know the theorem that the multiplicative group of a finite field is cyclic.

Here is a thread on the topic.

Finite subgroups of the multiplicative group of a field are cyclic

Edit:

The multiplicative group of the field with $q$ elements is therefore isomorphic to the additive group $\mathbf{Z}/(q-1)\mathbf{Z}$. There is an element of order 3 in this group if and only if $3 | (q-1)$.

Generally, if $n$ doesn't divide $q-1$, there will be no elements of order $n$ (but there might be some with order dividing $n$). If $n$ divides $q-1$, then $\mathbf{Z}/(q-1)\mathbf{Z}$ contains a unique subgroup of cardinal $n$, and there will be $\phi(n)$ elements of order $n$.

• I have already said that i am familiar with this result and I have tried to do to certain extent.. I would be thankful if you can help me to see how could this be useful... – user87543 Jan 5 '14 at 14:11
• I've edited the answer to give a further hint. – user119191 Jan 5 '14 at 14:14
• Oh yes.. I got it... So, number of cubes are $\frac{q-1}{3}$ if $3$ divides $q-1$ and $q-1$ if $3$ des not divide $q-1$ – user87543 Jan 5 '14 at 14:17

The number of elements that are $n$-th powers is $$\frac{q-1} {\gcd(n,q-1)}.$$ Here the denominator is the size of your kernel.

Lemma 1: $o(a^n)=\frac{o(a)}{\gcd(o(a),n)}$

Proof: Let $x=o(a), y=o(a^n), d=\gcd(x,n)$. We have $d|x$ and $d|n$, so $x=dv, n=du$. Then, $$\frac{o(a)}{\gcd(o(a),n)}=\frac{x}{d}=\frac{dv}{d}=v.$$ Since $x=o(a)$, we have $a^x=a^{dv}=1$ in $K$. Thus, $$(a^n)^v=(a^{du})^v=a^{duv}=(a^{dv})^u=1.$$ Thus, $y|v$. Then, $(a^n)^y=a^{ny}=1$, so $x|ny$. Thus, $$\frac{ny}{x}=\frac{duy}{dv}=\frac{uy}{v}\in\mathbf{Z}^+.$$ Now, by Bezout's Lemma, there exist $r,s$ such that $$rx+ns=d\implies rdv+dus=d\implies rv+us=1,$$ so $u$ and $v$ are coprime. Since $v|uy$ and $\gcd(u,v)=1$ we have $v|y$. However, we showed that $y|v$, so $v=y$. Finally, $x=dy$, so $o(a)\gcd(o(a),n)o(a^n)$, as desired. $\Box$

Suppose $K$ is a finite field of order $q$. Since $K^{\times}$ is cyclic, we can pick a generator $g$ in this group. Now, we have $$\text{number of dth powers}=\underbrace{\operatorname{ord}_{q}(g^d)=\frac{\operatorname{ord}_q(g)}{\gcd(\operatorname{ord}_q(g), d)}}_{\text{By Lemma 1}}=\frac{q-1}{\gcd(q-1, d)}.$$