Is it possible to a root of a Gaussian integer be a Hurwitz quaternion?

Question

I'm trying to solve the equation $$ E(x)=\frac{(1+ix)^{1/2}-(1-ix)^{1/2}}{(1+ix)^{1/2}+(1-ix)^{1/2}} \mod p, $$

where $p$ is a prime and $1+ix$ is a Gaussian integer over $p$. For some values of $x$, there is no solution for the $\sqrt{1+ix}$ in the complex field, i.e., is a quadratic nonresidue. I'm just an engineer student, my knowledge in the area is not great. There is a possibility that the solution is in an extension field like the Hurwitz quaternion?

Answer 1

What you propose is possible to do. Suppose $\,p\,$ is an odd prime. Let $$ E(x):=\frac{(1+ix)^{1/2}-(1-ix)^{1/2}}{(1+ix)^{1/2}+(1-ix)^{1/2}} \pmod p. $$ First, simplify the quotient to get (if $\,x\ne 0$ otherwise $E(0)=0$) $$ E(x) = i \frac{\sqrt{1+x^2}-1}x \pmod p. $$ Here we understand $\,i\,$ to mean a square root of $\,-1\pmod p\,$ if $\,p-1\,$ is a quadratic residue, or else it exists in an extension field. Again, the same problem arises for $\,\sqrt{1+x^2}\,$ if $\,1+x^2\,$ is not a quadratic residue and so use an extension field. Thus, we may need to use two extension fields to make sense of $\,E(x).\,$ Note that $\,E(-x)=-E(x)\,$ for all $\,x.\,$

For example, let $\,p=3.\,$ Define $\,i=\sqrt{-1}=\sqrt{2}\,$ in an extension field of $\,\mathbb F_3.\,$ We already know $\,E(0)=0.\,$ Now $\,E(1)=2+2i\,$ and $\,E(2)=1+i.\,$

Another example, let $\,p=5.\,$ Here $\,-1=2^2=3^2\pmod 5\,$ thus $\,i=2.\,$ Now we need $\,r:=\sqrt{2}\,$ in an extension field of $\,\mathbb F_5\,$ and thus, $$E(0)=0,\; E(1)=3+2r,\; E(2)=4,\; E(3)=1,\; E(4)=2+3r. $$

Answer 2

As @ThomasAndrews notes, your terminology is not quite standard, and gets tangled up with some mathematical issues, but perhaps I can guess what you are wanting.

Let $\mathbb F_q$ denote the finite field with $q$ elements. It is standard that there exists such a thing exactly for $q$ a power of a prime, and that then there is a unique ("up to isomorphism") such thing.

For $p$ a prime, $\mathbb F_p$ contains a $\sqrt{-1}$ if and only if $p=1\mod 4$ (or $p=2$, which is a degenerate case). So, for $p=3\mod 4$, adjoining a $\sqrt{-1}$ to $\mathbb F_p$ gives a strictly larger field $\mathbb F_{p^2}$. We could still call the $\sqrt{-1}$ "$i$", if we want.

I suspect that this is your set-up and you want to simplify the expression you wrote. So, yes, there is the subordinate question: for which $x\in \mathbb F_p$ is there a $\sqrt{1+ix}$ in $\mathbb F_{p^2}$ (for example).

An auxiliary fact (also standard), that may be useful to you, is that $x^2+1=y^2$ has a solution $(x,y)$ in $\mathbb F_p$, for every $p\not=2$. But, still, not every $x^2+1$ is a square in $\mathbb F_p$. Because of the uniqueness of $\mathbb F_{p^2}$, every $x^2+1$ is a square in that quadratic extension.

Some simplification can be done by "rationalizing the denominator", just as in more mundane algebra situations.

But/and if you can clarify what you're needing/wanting, more pointed answers could be given. :)

(As @ThomasAndrews already noted, there are no finite division algebra ("quaternions") except fields. But/and the theory of finite fields is simple and "happy" enough that many questions can be answered...)