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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?

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1 Answer 1

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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 $(\frac{1}{2}, \frac{1}{3}, \frac{1}{6})$.

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  • $\begingroup$ I think this looks good - think there is typo where you said $r_i = k_il_i$. $\endgroup$
    – dezdichado
    Mar 23, 2023 at 17:11
  • $\begingroup$ Thanks, yep it is supposed to be a fraction. $\endgroup$
    – C614
    Mar 23, 2023 at 17:15
  • $\begingroup$ Looks great, thanks! So the only solutions with no zeros are the ones that I found? $\endgroup$ Mar 26, 2023 at 6:29
  • $\begingroup$ Yes, Case 1, 3a, 3c are the only one that produce a solution. $\endgroup$
    – C614
    Mar 26, 2023 at 13:07

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