Is there any method to calculate the floored integer division for the sum of two numbers given the floored division of the summands, without splitting into cases?
I know that, with floored division, we have the two cases that if $a$%$b$ + $a$%$c$ $<c$ then $$\frac{a + b}{c} = \frac ac + \frac bc $$
and if $a$%$b$ + $a$%$c$ $\geq c$ then $$\frac{a + b}{c} = \frac ac + \frac bc + 1 $$
I wish to know if a non case-separated equation exists or not ?
No, you have to separate into cases; in particular, if we let $$f(x)=\left\lfloor\frac{x}c\right\rfloor$$ then this question asks if we can determine $f(a+b)$ given $f(a)$ and $f(b)$. As your cases show, we know that one of the following holds: $$f(a+b)=f(a)+f(b)$$ $$f(a+b)=f(a)+f(b)+1$$ However, to know which is which, we'd have to know something about the "excess" that we rounded off - that is, are $a-cf(a)$ and $b-cf(b)$ large enough to contribute an extra $1$ to $f(a+b)$? To be very explicit, notice that $n_1=f(cn_1)=f(c(n_1+1)-1)$ and $n_2=f(cn_2)=f(c(n_2+1)-1)$. If we could determine the sum $f(a+b)$ given just $f(a)$ and $f(b)$, then it would have to be that $f(cn_1+cn_2)$ equalled $f(c(n_1+1)-1+c(n_2+1)-1)$, since their summands have the same integer part in division. However, the former has $f(cn_1+cn_2)=n_1+n_2$ whereas the latter has $f(cn_1+cn_2+2c-2)=n_1+n_2+1$. Thus, the sum cannot be further determined just from $f(a)$ and $f(b)$.
We can, however, see that the identity: $$\left\lfloor\frac{a+b}{2c}\right\rfloor=\left\lfloor\frac{\left\lfloor\frac{a}c\right\rfloor+\left\lfloor\frac{b}c\right\rfloor}{2}\right\rfloor$$ does hold, since it holds in either case.