Sum of two floors is equal 
Let $a, b, c$, and $d$ be real numbers. Suppose that $\lfloor na\rfloor +\lfloor nb\rfloor =\lfloor nc\rfloor +\lfloor nd\rfloor $ for all positive integers $n$. Show that at least one of $a+b$, $a-c$, $a-d$ is an integer.


*

*My Progress:
Since $$1=\lim_{n\to \infty}{\frac{ \lfloor na\rfloor +\lfloor nb\rfloor }{\lfloor nc\rfloor +\lfloor nd\rfloor}}=\frac{a+b}{c+d}$$ Thus $a+b=c+d=x$.

*We have $\lfloor na\rfloor+\lfloor nx-na\rfloor=\lfloor nc\rfloor +\lfloor nx-nc\rfloor$. This means that $\{nx\}\geq \{na\} \equiv \{nx\}\geq\{nc\}$.
I think some clever bounding should do it but I am not able to do so...

 A: $n=1$ shows that one of $(a,b)$ is $\geq 1$ iff one of $(c, d)$ is $\geq 1$. Since we can add or subtract $1$ from one of $(a,b)$ and one of $(c,d)$ and the equation will still be satisfied, we may assume without loss of generality that $0 \leq a,b,c,d < 1$.
Write the base-$B$ representations of these numbers as $a = 0.a_{1,B}a_{2,B}a_{3,B}\cdots$ where each $a_{i,B}$ is a single digit,  and similarly for $b, c, d$. Using $n = B^k$ with induction on $k$, $a_{i,B} + b_{i,B} = c_{i,B} + d_{i,B}$ for every base $B$ and $i \geq 1$. [1]
For $B=2$ this shows that when the binary digits of $a$ and $b$ agree or disagree, the corresponding digits in $c$ and $d$ also agree or disagree.
If $a = b$ then this shows that $a=c$ and we are done.
Now assume $a \neq b$. Let $i_1, i_2, i_3, \ldots$ be the indices of binary digits on which $a$ and $b$ disagree (this may either be finitely long or an infinite sequence).
Let $A_k=0.a_{1,2}\cdots a_{k,2}$ be the number given by the binary representation of $a$ truncated to $k$ digits, and similarly for $b,c,d$.
Suppose by contradiction that $a \neq c$ and $a \neq d$. Then there must be some minimum $k$ such that $A_{i_k} \neq C_{i_k}$ and $A_{i_k} \neq D_{i_k}$, since $a$ must agree with $c$ and $d$ on all other binary digits. Since $c$ and $d$ also must disagree on binary digit $i_k$, $C_{i_k} \neq D_{i_k}$. Since $A_{i_k} + B_{i_k} = C_{i_k} + D_{i_k}$, $B_{i_k}$ also has a distinct value.
Let $e$ be the unique member of $\{a,b,c,d\}$ with maximum truncated value $E_{i_k}$. In base $Q=2^{i_k} E_{i_k}$, $E_{i_k}$ has representation $0.1000\cdots$. The remaining three of $A_{i_k},B_{i_k},C_{i_k},D_{i_k}$ have representations $0.0?????\cdots$ for some base-Q digits $?$. Since besides $e$ the other members of $a,b,c,d$ must have values less than $E_{i_k}$, it follows that $e$ has first base-Q digit $1$ while the others have first base-Q digit $0$, contradicting formula [1] for $B=Q$. Therefore either $a=c$ or $a=d$. QED
