Squares in a Finite Field

Question

Show that in any finite field,each of its elements can be written as the sum of two squares.

Well,I hate to admit-this being also my first post-that I have not proven it yet.I tried to work on the multiplicative group but to no avail.Any help/tip will be welcome.

Ah,I see its a repost basically..This is rather embarrassing,I should have looked more carefully.Thanks! — MathematicianByMistake, May 04 '15 at 14:00
The argument has been repeated many times on our site. In the one I declared this to be a duplicate of the emphasis is on prime fields, but both the answers give an argument that works for all finite fields. I picked that one because it is the oldest incarnation I found. — Jyrki Lahtonen, May 05 '15 at 05:19

score 16 · Accepted Answer · edited Dec 18 '19 at 20:53

16

Any finite field is of the form $\mathbb{F}_{p^k}$. If $p=2$ the cardinality of $\mathbb{F}_{p^k}^*$ is odd, hence every element of the multiplicative group is a square. So we may assume that $p$ is odd. In such a case, there are exactly $\frac{p^k-1}{2}$ squares in $\mathbb{F}_{p^k}^*$: let $A$ be their set and $Q$ be $A\cup\{0\}$.

For any $g\in\mathbb{F}_{p^k}$, both $Q$ and $g-Q$ have $\frac{p^k+1}{2}$ elements, hence they have to intersect (they cannot be disjoint). But: $$ q_1 = g-q_2,\quad q_i\in Q $$ gives: $$ g=q_1+q_2,\quad q_i\in Q $$ as wanted.

edited Dec 18 '19 at 20:53

Bach

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answered May 04 '15 at 14:07

Jack D'Aurizio

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Thank you Jack! Beautiful solution..To enhance my embarrassment,I did try to examine different cases depending on p,but not "dividing" the set.. – MathematicianByMistake May 04 '15 at 14:14
What exactly is your argument when $p = 2$? Also, what is $g - Q$? I think this answer needs some major reworking. – Junglemath Oct 15 '20 at 11:08
@Junglemath: the argument for $p=2$ is stated at the very beginning: in a multiplicative group with odd order every element is a square. And, as usual, $$ g-Q = {g-q: q\in Q}.$$ – Jack D'Aurizio Oct 16 '20 at 09:33
That seems like more of an assertion than an argument. – Junglemath Oct 16 '20 at 13:07
@Junglemath: still an obvious consequence of $g^{o(G)}=e$. – Jack D'Aurizio Oct 16 '20 at 16:33

Squares in a Finite Field

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