Math Help - Finite field

1. incidentally all of the above prove that any element can be written both as x^2+y^2 and x^2-y^2

2. Thank you very much for the long discussion and the good proof!

3. you are most welcome. i liked interacting with you. thank you too !!

4. Originally Posted by nirax
call $A$ as the multiplicative group of squares. so we decompose $F = A \cup \{0\} \cup B$. $B$ is the set of non squares. suppose an element $a$ is not expressible as sum of two squares. consider $a - A$. this set must be the same as $B$. consider $x = a - a^2$. clearly $x \in B$

but $B$ was the coset of the multiplicative group $A$ in $F^*$. so $B = a.A$. it follows that there exists a $y$ s.t

$x = a.y^2 = a - a^2$

$\Rightarrow a (y^2 - 1 + a) = 0$
$\Rightarrow y^2 = a - 1$

RHS belongs to $B$ whereas LHS belongs to $A$ which are disjoint.
hmm ....shouldn't it be y^2=1-a???

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