Number of elements of the field

[mathmari]

Hey!

Let $p$ an odd prime and $F=F_{p^n}$ the finite field with $p^n$ elements.

1. Show that the set $F^2=\{a^2, a \in F\}$ has $\frac{p^n+1}{2}$ elements. Conclude that , if $t \in F$ the set $t-F^2=\{t-a^2, a \in F\}$ has $\frac{p^n+1}{2}$ elements.
2. For $t \in F$ show taht the set $F^2 \cap (t-F^2)$ is non-empty and conclude that each element $c$ of $F$ can be written in the form $c=a^2+b^2, a, b \in F$.
3. Show that the equation $x^2+y^2+z^2=0$ has a non-trivial solution in $F$.
I have done the following:

1. $a^2 \in F^2 \Rightarrow a, -a \in F$

That means that $F^2$ contains the half of the elements of $F\setminus \{0\}$, and the element $0$.
So, the number of elements of $F^2$ is $\frac{p^n-1}{2}+1=\frac{p^n-1+2}{2}=\frac{p^n+1}{2}$.

Is this correct?? (Wondering)

How can I show that conclude that if $t \in F$ the set $t-F^2=\{t-a^2, a \in F\}$ has $\frac{p^n+1}{2}$ elements??

2. Let $x \in F^2 \cap (t-F^2) \Rightarrow x=F^2 \text{ AND } x \in t-F^2 \\ \Rightarrow x=a^2, a \in F \text{ AND } x=t-a^2, t, a \in F$

How can I continue?? (Wondering)

3. We set $z=1$, then we have the equation $x^2+y^2+1=0 \Rightarrow x^2=-1-y^2$.

Does $-1 \in F$ ?? (Wondering)

Then we could use the question $2.$, right?? (Wondering)

 

[Fallen Angel]

Hi,

1. Is correct

In 2. you haven't proved that the intersection is non-epmty, but it's just the pidgeonhole principle.

For the other question, if $F^{2}\cap (t-F^{2})$ is non empty, then there exists some $b,c \in F$ such that $b^{2}=t-c^{2}$, and this holds for every $t\in F$

In 3. Yes, $-1\in F$ becaus is the inverse with respect to the addition of the neutral element with respect to the multiplication.

And now by 2. you got it

 

[mathmari]

Hi,

1. Is correct

In 2. you haven't proved that the intersection is non-epmty, but it's just the pidgeonhole principle.

For the other question, if $F^{2}\cap (t-F^{2})$ is non empty, then there exists some $b,c \in F$ such that $b^{2}=t-c^{2}$, and this holds for every $t\in F$

In 3. Yes, $-1\in F$ becaus is the inverse with respect to the addition of the neutral element with respect to the multiplication.

And now by 2. you got it

1. How can I show that conclude that if $t \in F$ the set $t-F^2=\{t-a^2, a \in F\}$ has $\frac{p^n+1}{2}$ elements??

2. How can I show that the intersection is non-empty?? (Wondering)

3. We set $z=1$, then we have the equation $x^2+y^2+1=0 \Rightarrow x^2=-1-y^2$.

$-1 \in F$

By $2.$ we have that $c=a^2+b^2, a,b,c \in F \Rightarrow a^2=c-b^2$

For $a^2=x^2, c=-1, b^2=y^2$ we have that $x^2=-1-y^2$.

But how can we justify that the equation has a non-trivial solution in $F$ ?? (Wondering)

 

[Fallen Angel]

Hi,

1. The set $t-F^{2}$ it's just a traslation of $F^{2}$ so it has the same cardinality

2.I told you to use the pidgeonhole principle, you have two sets of $\frac{p^{n}+1}{2}$ elements contained in a set of $p^{n}$ elements, so ...

3.Since you have a solution with $z=1$ (by 2.) the solution is non trivial

 

[mathmari]

1. The set $t-F^{2}$ it's just a traslation of $F^{2}$ so it has the same cardinality

Ok! (Smile)

2.I told you to use the pidgeonhole principle, you have two sets of $\frac{p^{n}+1}{2}$ elements contained in a set of $p^{n}$ elements, so ...

Do you mean that $F^2 \subseteq F$ and $t-F^2 \subseteq F$ ?? (Wondering)

3.Since you have a solution with $z=1$ (by 2.) the solution is non trivial

And is the way I formulated it correct?? (Wondering)

We set $z=1$, then we have the equation $x^2+y^2+1=0 \Rightarrow x^2=−1−y^2$.

$−1 \in F$

By $2.$ we have that $c=a^2+b^2,a,b,c \in F⇒a^2=c−b^2$

For $a^2=x^2,c=−1,b^2=y^2$ we have that $x^2=−1−y^2$.

Or could I improve something?? (Wondering)

 

[johng1]

I'm a little dubious about your proof of 1. Where do you use the fact that p is odd? For p=2, $F^2=F$. So I think you need to expand your proof.

For 2, as was pointed out, if A is any subset of an additive group, then the cardinality of $t-A=\{t-a:a\in A\}$ is the same as the cardinality of A. (You can easily define a one to one function from A to t-A.) So suppose $(t-F^2)\cap F^2$ is empty. Then $p^{n+1}=2\,{p^n+1\over 2}=|t-F^2|+|F^2|=|(t-F^2)\cup F^2|\leq p^n$, a contradiction. (The last inequality since $(t-F^2)\cup F^2\subseteq F$)

 

[Fallen Angel]

Do you mean that $F^2 \subseteq F$ and $t-F^2 \subseteq F$ ?? (Wondering)

Yes

And is the way I formulated it correct?? (Wondering)

Or could I improve something?? (Wondering)

It's ok

 

[mathmari]

I'm a little dubious about your proof of 1. Where do you use the fact that p is odd? For p=2, $F^2=F$. So I think you need to expand your proof.

How could I expand the proof?? (Wondering)

For 2, as was pointed out, if A is any subset of an additive group, then the cardinality of $t-A=\{t-a:a\in A\}$ is the same as the cardinality of A. (You can easily define a one to one function from A to t-A.)

So, to show that $A$ and $t-A$ have the same cardinality do we have to show that this function is $1-1$ and onto?? (Wondering)

So suppose $(t-F^2)\cap F^2$ is empty. Then $p^{n+1}=2\,{p^n+1\over 2}=|t-F^2|+|F^2|=|(t-F^2)\cup F^2|\leq p^n$, a contradiction. (The last inequality since $(t-F^2)\cup F^2\subseteq F$)

When we suppose that $(t-F^2)\cap F^2$ is empty, does it mean that $|(t-F^2)\cap F^2|=|t-F^2|+|F^2|$ ?? (Wondering)

Also, why does it stand that $(t-F^2)\cup F^2\subseteq F$ ?? (Wondering)

 

[johng1]