Prove that if are $a, b, c, d$ real numbers for which holds $\lfloor na \rfloor+\lfloor nb \rfloor=\lfloor nc \rfloor+\lfloor nd \rfloor$ for all positive integers $n$ then $\{a+b\}\{a-c\}\{a-d\}=0$.
I was able to prove $a+b=c+d$ using the fact that $\frac{\lfloor an \rfloor+\lfloor bn \rfloor}{\lfloor cn \rfloor+\lfloor dn \rfloor}$ converges to $\frac{a+b}{c+d}$. I also have assumed that $a, b, c, d \in [0,1)$ which can be easely deduced from the relation.
When $a=0$ is quite easy to prove the statement. In my solution, I found a number $n$ for which $\{nd\}+\{nc\}>1$ which is impossible because $\{nb\}<1$. Anyways, my proof is pretty long so I am not going to write it, but there are also proofs for this on this forum. I will attach the link here:
I do not think this is going to help in the main proof because when $a$ is not $0$, this idea of construction is not really useful because I have 2 fractional parts in both of right and left hand.
Note: $\{x\}$ means fractional part of $x$.
My question is different from that with 3 fractional parts, which is easier. That question is a particular case of mine when one of the 4 numbers $a, b, c, d$ is 0. The general case is more complicated, so my question is not a DUBLICATE. If it so, then show me how, instead of closing my question. Prove me I'm wrong and show me how that old question will solve the general case, my question.