How prove this $\{a\}\cdot\{b\}\cdot\{c\}=0$ if $\lfloor na\rfloor+\lfloor nb\rfloor=\lfloor nc\rfloor$

Solution 1:

For $n=1$ one gets $\lfloor a\rfloor+\lfloor b\rfloor=\lfloor c\rfloor$. Subtracting $n$ times that equation from the given one gives $$ \lfloor n\{a\}\rfloor+\lfloor n\{b\}\rfloor=\lfloor n\{c\}\rfloor \quad\text{for all $n\in\Bbb N$,} $$ and together with $\lfloor a\rfloor+\lfloor b\rfloor=\lfloor c\rfloor$ these equations also imply the original ones. But now one only has the fractional parts $a'=\{a\}$, $b'=\{b\}$ and $c'=\{c\}$ in the formulation of the problem. This means it suffices to show the stated result in the special case $a,b,c\in[0,1)$. I will henceforth assume that.

First I show the conditions imply $a+b=c$. Suppose this equation fails, then choose $n\in\Bbb N$ such that $|a+b-c|\geq\frac2n$. From $|na+nb-nc|\geq2$ it follows that $\lfloor na\rfloor+\lfloor nb\rfloor\neq\lfloor nc\rfloor$, a contradiction.

It remains to prove that for $a,b\in(0,1)$ there always exists some $n\in\Bbb N$ such that $\{na\}+\{nb\}\geq1$, since that means $\lfloor na\rfloor+\lfloor nb\rfloor=nc-(\{na\}+\{nb\})=\lfloor nc\rfloor-1$ contradicting the hypothesis; assuming that, the cases of the reduced problem that satisfy the hypothesis must have $a=0\lor b=0$, which means that one has $\{a\}\cdot\{b\}=0$ in the original problem, stronger than what was needed.

Proving the existence of such an $n$ is a bit technical. If $a+b\geq1$ one can take $n=1$, so assume that $a+b<1$. Choosing $N>0$ with $\frac2N<\min(a,b,1-a-b)$, I claim there exists some $m$ such that $ma$ and $mb$ are both no further than $\frac1N$ from the nearest integer (call the actual differences $\epsilon_a,\epsilon_b$). Then one can take $n=m-1$ because $na\equiv -a+\epsilon_a\pmod1$ implies (due to the smallness of $\epsilon_a$) that $\{na\}=1-a+\epsilon_a$ and similarly $\{nb\}=1-b+\epsilon_b$; combining the two equations gives that $\{na\}+\{nb\}=2-a-b+\epsilon_a+\epsilon_b\geq1+(1-a-b-\frac2N)>1$.

The claim can be proved using the pigeonhole principle: the pairs of remainders modulo$~N$ of $\lfloor mNa\rfloor$ and of $\lfloor mNb\rfloor$ can take only finitely many (namely $N^2$) different values as $m$ varies, so there exist $m_1<m_2$ for which the same pair is obtained; then taking $m=m_2-m_1$, one has that the numbers $ma$ and $mb$ are both at distance less than $\frac1N$ from an integer.

Solution 2:

I will show a partial result. If $a,b,c$ are rational numbers then $a$ and $b$ cannot both be non-integers, i.e. we have $\{a\}\cdot \{b\}\cdot \{c\} = 0$.

We can wlog assume that $0 \leq a,b,c < 1$ since any integer-part can be subtracted away. We will also use OPs results that $a+b=c$. This results also implies

$$\{na\} + \{nb\} = \{nc\}$$

for all $n>0$. In the following we will assume that both $a$ and $b$ are rational numbers in $(0,1)$. Then there exist integers $k,N,p,M$ s.t.

$$a = \frac{k}{N}~~~~\text{and}~~~~b = \frac{p}{M}$$

with $k<N$, $p<M$, $(k,N)=1$ and $(p,M)=1$. First we note that $$\{c\} = \{a\} + \{b\} = \frac{k}{n} + \frac{p}{m}$$

By taking $n = NM-1$ and using the first equality above we get

$$\{na\} + \{nb\} = \{Mk-\frac{k}{N}\} + \{Np-\frac{p}{M}\} = 1 - \frac{k}{N} + 1- \frac{p}{M} = 2 - \{c\}$$

since $Mk,Np$ are integers and $0 < \frac{k}{N},\frac{p}{M} < 1$. It now follows that

$$\{nc\} + \{c\} = 2$$

which is impossible since $\{c\},\{nc\} < 1$. There is one last case that is not covered by the argument above and that is if $a+b = 1$. This case is however ruled out by $\{a\} + \{b\} = 1 \not= \{c\} = 0$.

Solution 3:

Using Winther result I can finish the proof for other cases. In order to do this we have to deduce this problem to rationals.

According to little bit stricter version of Dirichlet's approximation theorem we can find infinitely many pairs of fraction $\left(\frac{p_1}q, \frac{p_2}q\right)$ with the following property:

$$\max\left(\left|a - \frac{p_1}q\right|, \left|b - \frac{p_2}q\right|\right) < q^{-\frac32}.$$

Infinitely many means that we can make $q$ as large as we want, we will choose it later. Finally, let's take $n = q-1$:

$$ \{n\frac{p_1}{q}\} + \{n\frac{p_2}{q}\} = \{1-\frac{p_1}{q}\} + \{1-\frac{p_2}{q}\} = 2 - \frac{p_1}{q}-\frac{p_2}{q} =\\ = 2 - \{c\} + \left(a - \frac{p_1}{q}\right) + \left(b - \frac{p_2}{q}\right) = 2 - \{c\}+\frac{\alpha}{q}, \alpha \in [-2, 2] $$

Also from Dirichlet's theorem we know that $n\left(a - \frac{p_1}{q}\right) < q^{-\frac12}$ and the same thing with $b$. Finally we can get:

$$ 2 - \alpha q^{-\frac12} = \{nc\}+\{c\} \leq 1 + \{c\}, \alpha \in [-4, 4] $$

Now we can take very large $q$ and get contradiction.

Edit. In fact I omitted one little issue, namely, why the following implication is true?

$$ (na - n\frac{p_1}q) < q^{-\frac12} \Rightarrow (\{na\} - \{n\frac{p_1}q\}) < q^{-\frac12} $$

Unfortunately, it is not true in every single case, but in our situation we can handle it. Remember that $n = q-1$. The problem can occur if there is an $A \in \mathbb{Z}$ (in fact $A = p_1$) such that

$$ n\frac{p_1}q=(q-1)\frac{p_1}q = p_1 - \frac{p_1}q < A < (q-1)a = na $$

And also we know that distance between $n\frac{p_1}q$ and $na$ less than $q^{-\frac12}$. Last thing that we should mention is that $\frac{p_1}q > \frac{a}2 > 0$ since $\frac{p_1}q$ is close to $a \neq 0$.