While working on a problem I ended up with the form
$$ 4n = u^2 + w^2 - (a+p)^2 - (b-r)^2 $$
or more generally,
$$ 4n = a_1^2 + a_2^2 - a_3^2 - a_4^2. $$
Question: Does this form represent all positive integer multiples of $4$?
In [Dickson1927], I found this theorem which seems to mention universal forms for $4n$.
I am not sure if we can derive the form given in this question from this theorem.
Update (Oct 24, 2022): If we take $a-p = u, ap = v, b+r = w, br = x$ and if $n = ap + br = v + x$ then we have
$$\begin{align} a_1 & = a-p = u, \\ a_2 & = b+r = w, \\ a_3 & = a+p, \\ a_4 & = b-r. \end{align}$$
As outlined in this question, factor $n = UV$ where $U,V,g \in \mathbb{Z}$ not necessarily prime and write
$$ n = UV = \overbrace{(U + V - g)}^a \overbrace{(g)}^p + \overbrace{(U - g)}^b \overbrace{(V-g)}^r $$
We can see from the equation above that $U = p + b$ and $V = p + r$.
Since all integers may be written as $n = UV$ where $U,V \in \mathbb{Z}$, we have
$$ 4n = a_1^2 + a_2^2 - a_3^2 - a_4^2. $$
Therefore, this form universally represents multiples of $4$.
References:
[Dickson1927]: Dickson, L. E. “Quaternary Quadratic Forms Representing All Integers.” American Journal of Mathematics, vol. 49, no. 1, 1927, pp. 39–56. JSTOR, https://doi.org/10.2307/2370770. Accessed 23 Oct. 2022.