Under $ad-bc=1$, is every element of a finite field of the form $a^2+b^2+c^2+d^2$?

Question

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).

Answer 1

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

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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 $$ \left\{ \begin{array}{rcl} a&=&\frac{s+t}2,\\ b&=&\frac{u+v}2,\\ c&=&\frac{u-v}2,\\ d&=&\frac{s-t}2. \end{array}\right. $$

Answer 2

We first treat the case where $\mathbb F_q$ has characteristic two. If $x=0$, we may take $a=d=1$ and $b=c=0$. If $x\neq 0$, then there exists some square root $a$ of $x$; note $a\neq 0$. Taking $b=0$ and $c=d=1/a$, we have $$ad-bc=a(1/a)-0=1;\qquad a^2+b^2+c^2+d^2=x+0+(1/a)^2+(1/a)^2=x.$$

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We now treat the odd characteristic case. We count the number $N$ of solutions to $a^2+b^2+c^2+d^2=ad-bc=0$. Inspired by the comment of Amateur_Algebraist, whenever this is nonzero modulo the characteristic $p$, Theorem 2.2(a) of this article:

Theorem 2.2 (Chevalley–Warning at the Boundary, Preliminary Form). Let $f_1, \ldots, f_r \in \Bbb F_q[t_1, \ldots, t_n]$ be polynomials of degrees $d_1, \ldots, d_r \in \Bbb Z^+$, and suppose that $d:= \sum_{j=1}^r d_j \le n.$ Let $E: \Bbb F^n_q \to \Bbb F^r_q$, $x \mapsto (f_1(x), \ldots, f_r(x))$ be the associated evaluation map. Then: (a) For all $b,c \in \Bbb F_q^r$ we have $ |E^{-1}(b)| \equiv |E^{-1}(c)| (\text{mod } p)$.

will imply that the desired equation has a solution. We will be able to explicitly compute $N$. It is useful to write $T$ for the number of solutions to $x^2+y^2=0$ in $\mathbb F_q$ with $x$ and $y$ nonzero.

1. If one of the variables are zero, another must (by $ad-bc=0$), while if three of them are zero then all four must be (by $a^2+b^2+c^2+d^2=0$). There are four ways to choose exactly two of the variables to be zero, as we must choose one from $\{a,d\}$ and one from $\{b,c\}$. Regardless of which pair we choose, the number of solutions in this case is equal to the number of solutions to $x^2+y^2=0$ with $x,y\neq 0$, i.e. $T$. We conclude that the number of solutions to our equations with some variables set to zero is $$1+4T.$$

2. If none of the terms are zero, the equation $ad-bc=0$ allows us to write $a=wx$, $d=yz$, $b=wy$, $c=xz$ for $w,x,y,z\in\mathbb F_q^\times$; these are unique up to scaling $w$ and $z$ by some scalar $s$ and scaling $x$ and $y$ by $s^{-1}$. So, the number of solutions to $a^2+b^2+c^2+d^2=0$ with $ad-bc=0$ and $0\not\in\{a,b,c,d\}$ is $$\frac1{q-1}\big(\text{# of solutions to }(wx)^2+(wy)^2+(xz)^2+(yz)^2=0\text{ in }w,x,y,z\in\mathbb F_q^\times\big).$$ This expression factors as $(w^2+z^2)(x^2+y^2)$. So, it is zero if and only if at least one of the factors is zero. The number of ways to make both factors nonzero is $(q-1)^2-T$, and so the number of solutions is $(q-1)^4-((q-1)^2-T)^2=2T(q-1)^2-T^2$. We conclude that the number of solutions to our equations with no variables equal to zero is $$2T(q-1)-\frac1{q-1}T^2.$$

Now, we evaluate $T$. If $-1$ is not a square in $\mathbb F_q$, then $T=0$. This implies that $N=1$. If $-1$ is a square in $\mathbb F_q$, then (since $p\neq 2$) each $x$ gives two solutions $y$, and so $T=2(q-1)$. This gives $$N=1+4T+2T(q-1)-\frac1{q-1}T^2=1+8(q-1)+4(q-1)^2-4(q-1)=(2q-1)^2.$$ In either case, $N\equiv 1\pmod p$.