Abstract Algebra – Under ad-bc=1, Is Every Element of a Finite Field of the Form a^2+b^2+c^2+d^2?

abstract-algebrafield-theoryfinite-fields

The Question:

Let $x\in\Bbb F_q$, where $\Bbb F_q$ is the field of $q=p^r$ elements, $p$ prime, $r\in\Bbb N$. Can we write

$$x=a^2+b^2+c^2+d^2\tag{1}$$

for $a,b,c,d\in\Bbb F_q$ such that $ad-bc=1$?

Thoughts:

Relevant theorems include:

  • Lagrange's Four Squares Theorem: Each natural number is the sum of four squares.
  • The number of squares in $\Bbb F_q$ is $q$ if $q$ is even, and $\frac{q+1}{2}$ if $q$ is odd.
  • Each element of a finite field is the sum of two squares.

I have tested via GAP that it holds for prime powers up to $q=101$.

Motivation:

I was playing around with traces of elements of $\operatorname{SL}_2(\Bbb F_q)$.

I have intuitive, hard-to-articulate reasons to suspect the answer to my question is "yes".

It would help my research to know an answer; however, field theory is not my forte. I will, of course, credit whoever answers first (if this is a new problem).

Best Answer

Carl's answer is nice. I am adding an elementary argument.

If $q$ is even, then all the elements are squares, and the question is trivial as we can use $a=d=1, b=0$. So assume that $q$ is odd, when $1/2$ exists as an element of $\Bbb{F}_q$.

We can then diagonalize the quadratic form yielding the determinant: $$ ad-bc=\frac14\left[(a+d)^2-(a-d)^2\right]+\frac14\left[(b-c)^2-(b+c)^2\right]. $$ It is then easy to see that both equations can be written in terms of the new variables $a\pm d$, $b\pm c$, and we want to prove the solvability of the system $$ \left\{ \begin{array}{ccl} (a+d)^2-(a-d)^2-(b+c)^2+(b-c)^2&=&4,\\ (a+d)^2+(a-d)^2+(b+c)^2+(b-c)^2&=&2x, \end{array}\right.\qquad(*) $$ where $x$ is the element to be represented as $a^2+b^2+c^2+d^2$. Writing $s=a+d$, $t=a-d$, $u=b+c$, $v=b-c$, we see that the system $(*)$ is satisfied if and only if $$ s^2+v^2=x+2\qquad\text{and}\qquad t^2+u^2=x-2.\qquad (**) $$ The OP mentioned the well known fact that every element of a finite field can be written as a sum of two squares. This implies the existence of a pair $s,v\in\Bbb{F}_q$ such that $s^2+v^2$ has the prescribed value $x+2$. Similarly, we see the existence of a pair of elements $t,u\in\Bbb{F}_q$ such that $t^2+u^2=x-2$.

But, again because we are in odd characteristic, we can solve $$ \left\{ \begin{array}{rcl} a&=&\frac{s+t}2,\\ b&=&\frac{u+v}2,\\ c&=&\frac{u-v}2,\\ d&=&\frac{s-t}2. \end{array}\right. $$

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