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i need help with this exercise.

If $\mathbb{H}$ is a finite field and $a\neq0, b\neq0$, are elements of $\mathbb{H}$ then exists $u,v\in\mathbb{H}$ such that $1+au^{2}+bv^{2}=0 $

I don't have idea of how attack this exercise. Can someone help me?

rcoder
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    Just a comment. The set $K$ of all squares in $H-{0}$ is a subgroup, usually of index $2$. So the set of elements $au^2$ for $u \ne 0$ is either $K$ or its only coset. Not sure if this helps. – user49640 Jun 14 '17 at 23:34
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    if your field has $p^n$ elements with $p$ odd then there are $\frac {p^n+1}2$ elements of the form $-bv^2$ and $\frac {p^n+1}2$ elements of the form $1+au^2$ . As usual, characteristic $2$ needs special arguing. – lulu Jun 14 '17 at 23:42
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    For characteristic 2, put $v=0$ and $u = \sqrt{a^{-1}}$ – J. Doe Jun 15 '17 at 02:16
  • The trick described by Lulu has been explained many times on our site. The one I selected as the duplicate target was just the highest voted version that turned up with the search buzzwords I used this time. – Jyrki Lahtonen Jun 15 '17 at 06:13

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