Recall that if $A$ and $B$ are both subsets of the integers, then $A+B=\{a+b:a \in A,b \in B\}$. Lagrange's four-square theorem states that if $A$ is the set of squares, then $4A=A+A+A+A=\mathbb{N}$.
Suppose we fix some integer $m \in \mathbb{N}$, and some residue $a \mod m$. By Lagrange's four-square theorem, all positive integers $n \equiv a \mod m$ can be written as a sum of four squares. Let $$E=\{(x_1,x_2,x_3,x_4) \in (\mathbb{Z}/m\mathbb{Z})^4: x_1^2+x_2^2+x_3^2+x_4^2 \equiv a \mod m\}.$$
Fix some element $(x_1,x_2,x_3,x_4) \in E$, and for each $i \in \{1,2,3,4\}$, let $A_i=\{k^2 \in \mathbb{N}: k \equiv x_i \mod m\}$.
Is every sufficiently large positive integer $n \equiv a \mod m$ in $A_1+A_2+A_3+A_4$? If this is the case, how large should we expect $n$ to be to guarantee this?
Let $$r_4(n)=|\{(y_1,y_2,y_3,y_4) \in \mathbb{Z}^4: y_1^2+y_2^2+y_3^2+y_4^2=n\}|,$$ and let $$r_4(n;x_1,x_2,x_3,x_4)=|\{(y_1,y_2,y_3,y_4) \in \mathbb{Z}^4: y_1^2+y_2^2+y_3^2+y_4^2=n,y_i \equiv x_i \mod m \text{ for each i}\}|$$ Assuming that every sufficiently large positive integer $n \equiv a \mod m$ is in $A_1+A_2+A_3+A_4$, should we expect some sort of equidistribution with respect to this property? What I mean by this is: For a sufficiently large positive integer $n \equiv a \mod m$, and every $(x_1,x_2,x_3,x_4) \in E$, is it true that $$r_4(n;x_1,x_2,x_3,x_4) \approx \frac{r_4(n)}{|E|}?$$ Perhaps, if $n$ is not square-free, one can force some grouping of the solutions, but what about if we add the condition that $n$ be square-free? On average, one can verify without too much effort that this is true, that is that $$\sum_{n<x,n \equiv a \mod m} r_4(n;x_1,x_2,x_3,x_4) \sim \frac{1}{|E|} \sum_{n < x, n\equiv a \mod m} r_4(n),$$ but it is unclear to me if it is true in general, not just on average.
