Let $x,y \in \mathbb{Z}$, and suppose that $x^2+y^2 \ge 4$. Do there exist $a,b,c,d \in \mathbb{Z}$ such that $ (a+d)^2+(b-c)^2=x^2+y^2 $ and $ad-bc=1$?
This question is motivated by an attempt to characterise the norms of matrices in $SL_2(\mathbb{Z})$: If $A =\begin{pmatrix} a & b \\\ c & d \end{pmatrix} \in SL_2(\mathbb{Z})$, then $$ (a+d)^2+(b-c)^2=\|A\|^2+2, $$ so I wonder whether $\|A\|^2+2$ can be any sum of squares.
Let $x=1$ and $y=3$ so that $x^2+y^2=10$. Then if $\tbinom{a\ b}{c\ d}\in\operatorname{SL}_2(\Bbb{Z})$ is such that $$(a+d)^2+(b-c)^2=x^2+y^2=10,\tag{1}$$ then $\{a+d,b-c\}=\{\pm 1,\pm 3\}$ for some for some choices of the signs. Then of course the matrices $$\begin{pmatrix}-a&\hphantom{-}b\\\hphantom{-}c&-d\end{pmatrix}, \begin{pmatrix}\hphantom{-}a&-b\\-c&\hphantom{-}d\end{pmatrix}, \begin{pmatrix}\hphantom{-}b&\hphantom{-}a\\-d&-c\end{pmatrix} \in\operatorname{SL}_2(\Bbb{Z}),$$ also satisfy $(1)$, so without loss of generality $a+d=1$ and $b-c=3$. Then $$1=ad-bc=a(1-a)-b(b-3),$$ which shows that $b$ is an integral root of a quadratic with discriminant $$\Delta=-4a^2+4a+5.$$ But this is never a perfect square; a contadiction! Hence no such matrix exists.