Theorem: Seventeen Horses/General Problem 1

A man dies, leaving $n$ indivisible and indistinguishable objects to be divided among $3$ heirs.
They are to be distributed in the ratio $\dfrac 1 a : \dfrac 1 b : \dfrac 1 c$.
Let $\dfrac 1 a + \dfrac 1 b + \dfrac 1 c < 1$.
Then there are $7$ possible values of $\tuple {n, a, b, c}$ such that the required shares are:
:$\dfrac {n + 1} a, \dfrac {n + 1} b, \dfrac {n + 1} c$
These values are:
:$\tuple {7, 2, 4, 8}, \tuple {11, 2, 4, 6}, \tuple {11, 2, 3, 12}, \tuple {17, 2, 3, 9}, \tuple {19, 2, 4, 5}, \tuple {23, 2, 3, 8}, \tuple {41, 2, 3, 7}$
leading to shares, respectively, of:
:$\tuple {4, 2, 1}, \tuple {6, 3, 2}, \tuple {6, 4, 1}, \tuple {9, 6, 2}, \tuple {10, 5, 4}, \tuple {12, 8, 3}, \tuple {21, 14, 6}$

Proof:
It is taken as a condition that $a \ne b \ne c \ne a$.
We have that:
:$\dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 n = 1$
and so we need to investigate the solutions to the above equations.
From Sum of 4 Unit Fractions that equals 1, we have that the only possible solutions are:
{{begin-eqn}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6
      | r = 1
}}
{{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}
      | r = 1
}}
{{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}
      | r = 1
}}
{{eqn | l = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6
      | r = 1
}}
{{eqn | l = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4
      | r = 1
}}
{{end-eqn}}
From these, we can eliminate the following, because it is not the case that $a \ne b \ne c \ne a$:
{{begin-eqn}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6
      | r = 1
}}
{{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}
      | r = 1
}}
{{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}
      | r = 1
}}
{{eqn | l = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6
      | r = 1
}}
{{eqn | l = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4
      | r = 1
}}
{{end-eqn}}
Then we can see by inspection that the following are indeed solutions to the problem:
{{begin-eqn}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}
      | r = 1
}}
{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8
      | r = 1
}}
{{end-eqn}}
The remaining tuple we have is:
:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15} = 1$
But we note that:
:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} = \dfrac {28} {30}$
which is not in the correct form.
Hence the $7$ possible solutions given.
{{qed}}