Let $F$ be a field. Consider the following three abilities:
- (SQRT) Given $a\in F$, find $x\in F$ such that $a = x^2$ (when such $x$ exists).
- (NORM) Given $a,b\in F$, find $x\in F$ such that $a^2+b^2 = x^2$ (when such $x$ exists).
- (BISQ) Given $a\in F$, find $x,y\in F$ such that $a = x^2+y^2$ (when such $x,y$ exist).
Abilities NORM and BISQ together yield ability SQRT. Ability SQRT (plus the ability to perform field operations) yields ability NORM. If $F=\mathbb{R}$, then ability SQRT also yields ability BISQ (since in $\mathbb{R}$ the numbers expressible as a sum of squares are exactly the numbers expressible as squares, namely the nonnegative numbers, so we can compute $x$ using SQRT and take $y=0$), and so for $\mathbb{R}$ ability SQRT is equivalent to the combination of abilities NORM and BISQ. [This paragraph was revised after an error was pointed out by joriki.]
My question: Is NORM a strictly lesser capability than SQRT (in $\mathbb{R}$ or in any other field)?
I don't state the question formally because I don't know the right formalization. Any reasonable interpretation of the question is legit.
My observations so far:
In a field of characteristic 2, ability NORM is trivial (take $x=a+b$). It seems like SQRT should be a nontrivial ability, but I don't see how to prove that, for example, in the field of power series over $\mathbb{Z}/2\mathbb{Z}$, the square root operation cannot be performed using the field operations alone.
I had hoped to construct a field $F$ containing $\mathbb{Q}$ and closed under the operation of NORM, but not closed under SQRT (maybe using some trick related to the characterization of the integers which are representable as the sum of two squares). This would show that, in that field, NORM is a strictly lesser capability than SQRT in a very strong sense. But because every positive integer is the sum of four squares, any such field contains the square roots of all positive integers (if $n = a^2+b^2+c^2+d^2$ then $\sqrt n = \sqrt{(\sqrt{a^2+b^2})^2 + (\sqrt{c^2+d^2})^2}$), hence the square roots of all positive rationals ($\sqrt{p/q} = \sqrt{pq}/q$), so it doesn't seem very hopeful.
(The origin of the question is that a computer programmer I know was wondering whether there was a way to perform NORM without having to invoke the computationally expensive SQRT operation. I doubt that this question has a positive practical answer, but the theoretical aspect struck my curiosity.)