a^2+b^2 in R^x iff (a,b) in S^x show $\ a^2+b^2\in R^x$ iff $\ (a,b)\in S^x$
Here is my thinking so far, since we showed that $\ (a,b)\in S^x$ iff $\ (a,-b) \in S^x$ this means the product  $\ (a,b)(a,-b) = (a^2 + b^2, 0)$ is invertible. So does this imply that the first component is invertible in R? Also if this is correct how would I go about showing the other direction? 
Thank you.

If $(u,v)\in S$ is the inverse of $(a,b)$, then $(u,-v)$ is the inverse of $(a,-b)$.
Consequently,
\begin{align}
(1,0)
&=(a,b)(u,v)(a,-b)(u,-v)\\
&=(a^2+b^2,0)(u^2+v^2,0)\\
&=((a^2+b^2)(u^2+v^2),0)
\end{align}
hence $(a^2+b^2)(u^2+v^2)=1$ in $R$ hence $a^2+b^2\in R^\times$.
Conversely, if $a^2+b^2$ is invertible in $R$, then $(a^2+b^2)t=1$ for some $t\in R$.
Consequently,
\begin{align}
(1,0)
&=(a^2+b^2,0)(t,0)\\
&=(a,b)(a,-b)(t,0)
\end{align}
so that $(a,b)$ is invertible in $S$ with inverse $(a,-b)(t,0)$.

For another point of view, note that $S\cong R[X]/\langle X^2+1\rangle$.
Putting $i=(0,1)\in S$, we can assume $S=R[i]$ where $i^2=-1$.
We have an $R$-algebra homomorphism $\sigma:S\to S$ satisfying $\sigma(i)=-i$.
Then
\begin{align}
N&:S\to R&
&s\in S\mapsto s\sigma(s)
\end{align}
is a (multiplicative) monoid homomorphism satisfying $N(a+ib)=a^2+b^2$ for every $a,b\in R$.
Then $N$ induces an homomorphism between groups of units $S^\times\to R^\times$.
Consequently, $a+ib\in S^\times$ implies $a^2+b^2\in R^\times$.
Conversely, if $s=a+ib\in S$ and $N(s)\in R^\times$, then $s\sigma(s)N(s)^{-1}=1$ in $S$, hence $s\in S^\times$ with inverse $s^{-1}=\sigma(s)N(s)^{-1}$.
