Below, $r_1,r_2,r_3$ are positive real numbers such that $r_1+r_2+r_3=1$, $m$ is an arbitrary positive integer, and $m_i := \lfloor r_i\cdot m\rfloor$ for all $i\in [3]$.
Define the triplet $(r_1,r_2,r_3)$ as good if the following holds for every positive integer $m$:
If
$$m_1 + m_2 + m_3 = m-2,$$
then
$$\frac{m_1+1}{r_1} = \frac{m_2+1}{r_2} = \frac{m_3+1}{r_3} = m+1$$.
So far, I found only three good triplets (up to permutations):
1. $(1/3, 1/3, 1/3)$ is good. Proof: If $r_i=1/3$ for all $i\in [3]$, then $m_i = \text{floor}(m/3)$. The condition $m_1 + m_2 + m_3 = m-2$ implies that the sum of fractions of $r_1 m$, $r_2 m$ and $r_3 m$ equals $2$, so each of these fractions must be $2/3$. This means that $m_i = m/3-2/3$, so $m_i+1 = m/3+1/3$, so $(m_i+1)/r_i = m+1$.
2. $(1/2, 1/4, 1/4)$ is good: Proof: again, the condition $m_1 + m_2 + m_3 = m-2$ implies that the sum of fractions of $r_1 m$, $r_2 m$ and $r_3 m$ equals $2$. This is possible only when $m \mod 4 = 3$, when the fractions are $1/2$, $3/4$ and $3/4$ respectively. So $m_1+1 = m/2+1/2$ and $m_2+1=m_3+1=m/4+1/4$. Hence, $(m_i+1)/r_i = m+1$.
3. $(1/2, 1/3, 1/6)$ is good. The proof is similar to case 2 above.
Are there any other good triplets? Or, is it possible to prove that these are the only good triplets?
Best Answer
You can start with noticing that triplets with at least one zero work automatically, since the assumption $m_1+m_2+m_3=m-2$ is never satisfied.
I will also assume that triplet $r$ is rational, since the conclusion $\frac{m_i+1}{r_i}=m+1$ will never hold for irrational $r_i$, while the assumptions $m_1+m_2+m_3=m-2$ will sometimes hold.
So assume that triplet is rational and for $i\in\{1,2,3\}$ let $r_i=\frac{k_i}{\ell_i}$ where $k_i$ and $\ell_i$ are relatively prime positive integers. There are a few cases to consider. Basic idea is to consider values of $m$ around $\ell_1\ell_2\ell_3$.
Case 1$\frac{k_i}{\ell_i}<\frac{1}{2}$ for all $i$. Consider $m=\ell_1\ell_2\ell_3+2$. Then $m_1=k_1\ell_2\ell_3$, $m_2=k_2\ell_1\ell_3$, $m_3=k_3\ell_1\ell_2$, and so $m_1+m_2+m_3=m-2$. Then the condition $\frac{m_1+1}{r_1}=m+1$ becomes $\ell_1\ell_2\ell_3+\frac{\ell_1}{k_1}=\ell_1\ell_2\ell_3+3$, and so $\frac{k_1}{\ell_1}=\frac{1}{3}$. Similiarly other fractions have to be $\frac{1}{3}$.
Case 2 for some $i$ we have $\frac{k_i}{\ell_i}>\frac{1}{2}$. Let this be for $i=1$, so $2k_1>\ell_1$. Consider $m=\ell_1\ell_2\ell_3-2$. Then $m_1=k_1\ell_2\ell_3-2$, $m_2=k_2\ell_1\ell_3-1$, $m_3=k_3\ell_1\ell_2-1$, and so $m_1+m_2+m_3=m-2$. Then the condition $\frac{m_1+1}{r_1}=m+1$ becomes $\ell_1\ell_2\ell_3-\frac{\ell_1}{k_1}=\ell_1\ell_2\ell_3-2$, but left side is not an integer, while the right side is. So no solutions in this case.
Case 3 $r_i=\frac{1}{2}$ for some $i$. Assume $r_1=\frac{1}{2}$. There are three subcases here as well.
Case 3a both $r_2<\frac{1}{3},r_3<\frac{1}{3}$. Then consider $m=2\ell_2\ell_3+3$. Then $m_1=\ell_2\ell_3+1$, $m_2=2k_2\ell_3$, $m_3=2k_3\ell_2$. Condition $\frac{m_2+1}{r_2}=m+1$ implies that $\frac{\ell_2}{k_2}=4$, or that $r_2=\frac{1}{4}$. Similarly, $r_3=\frac{1}{4}$ and you get your solution $(\frac{1}{2}, \frac{1}{4}, \frac{1}{4})$.
Case 3b one of $r_2,r_3$ is greater than $\frac{1}{3}$. Assume $r_2>\frac{1}{3}$. Then consider $m=2\ell_2\ell_3-3$. Then $m_1=\ell_2\ell_3-2$, $m_2=2k_2\ell_3-2$, $m_3=2k_3\ell_2-1$. Condition $\frac{m_2+1}{r_2}=m+1$ implies that $\frac{\ell_2}{k_2}=2$, or that $r_2=\frac{1}{2}$, which implies $r_3=0$.
Case 3c One of $r_2,r_3$ is $\frac{1}{3}$. This leads to your solution $(\frac{1}{2}, \frac{1}{3}, \frac{1}{6})$.