<s>[INST] Expression of sum of squares as a sum of a specific form This question is a follow-up of this one.
Let $x,y \in \mathbb{Z}$, and suppose that $x^2+y^2 \ge 4$, and that $x,y$ are not both odd.
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 $x^2+y^2$ where $x,y$ are not both odd.
This parity restriction on $x,y$ is certainly necessary for the existence of such a representation:
If $x,y$ are both odd, then $a+d,b-c$ must be odd (by looking at equality $\mod 4$). Thus one of $a$ and $d$, and one of $b$ and $c$ must be even, hence $ad-bc$ is even which is incompatible with $ad-bc=1$.
 [/INST] Let $x=0$ and $y=3$ so that $x^2+y^2=9$.  Then if $\tbinom{a\ b}{c\ d}\in\operatorname{SL}_2(\Bbb{Z})$ is such that
$$(a+d)^2+(b-c)^2=x^2+y^2=9,\tag{1}$$
then $\{a+d,b-c\}=\{0,\pm 3\}$ 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 $a+d=0$ and $b-c=3$. Then
$$1=ad-bc=a(-a)-b(b-3),$$
which shows that $b$ is an integral root of a quadratic with discriminant
$$\Delta=-4a^2+77.$$
But this is never a perfect square; a contadiction! Hence no such matrix exists.
</s>