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Lemma. If $F$ is a finite field, $\alpha\neq0\in F$ then there exists $\lambda,\,\mu\in F$ so that $1+\lambda^{2}-\alpha\mu^{2}=0$.

Proof. If the characteristic of $F$ is $2$, $F$ has $2^n$ elements and every element $x$ in $F$ satisfies $x^{2^n}=x$. This every element in $F$ is a square. In particular $\alpha^{-1}=\mu^{2}$. Using this $\mu$ and $\lambda=0$ we get $1+\lambda^{2}-\alpha\mu^{2}=1+0-1=0$.

My problems starts when the characteristic is an odd prime $p$, $F$ has $p^n$ elements. Could you help me to continue this proof, please?

Could you also check the first part of these proof?

Thank you.

Monika
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1 Answers1

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Consider the set $S_1=\{1+\lambda^2 : \lambda\in F\}$ and $S_2=\{a\mu^2 : \mu\in F \}$. How many elements are in $S_1$? Well, if $1+\lambda_1^2=1+\lambda_2^2$, then $\lambda_1=\pm \lambda_2$. So if $\lambda_1\neq 0$, then $\lambda_1$ and $-\lambda_1$ contribute one element to $S_1$. Counting $0$ separately, we get that $$|S_1|=1+\frac{p^n-1}{2}=\frac{p^n+1}{2}$$ Using the same method, one also gets that $$|S_2|=\frac{p^n+1}{2}$$ But now, $$|S_1|+|S_2|=\frac{p^n+1}{2}+\frac{p^n+1}{2}=p^n+1>p^n=|F|$$ Hence, there exists some $x\in S_1\cap S_2$. But then $x=1+\lambda^2$ for some $\lambda\in F$ and $x=a\mu^2$ for some $\mu\in F$. It is now seen that $1+\lambda^2-a\mu^2=0$, as desired.

And yes, your proof for characteristic 2 is correct.

Prism
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