Let: :$1 = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 d$ where: :$a \le b \le c \le d$ and: :$a \ge 2$ === $a = 2$ === Let $a = 2$. ==== $b = 2$ ==== Let $b = 2$. Then: :$\dfrac 1 a + \dfrac 1 b = 1$ leaving no room for $c$ and $d$. Hence there are no solutions where $a = 2$ and $b = 2$. {{qed|lemma}} ==== $b = 3$ ==== Let $b = 3$. {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b       | r = \dfrac 1 2 + \dfrac 1 3       | c =  }} {{eqn | r = \dfrac {3 + 2} 6       | c =  }} {{eqn | r = \dfrac 5 6       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c + \dfrac 1 d       | r = 1 - \dfrac 5 6       | c =  }} {{eqn | r = \dfrac 1 6       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c       | o = <       | r = \dfrac 1 6       | c =  }} {{eqn | ll= \leadsto       | l = c       | o = >       | r = 6       | c =  }} {{end-eqn}} Thus we try $c = 7, 8, \ldots$ in turn. $c = 7$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7       | c =  }} {{eqn | r = \dfrac {21 + 14 + 6} {42}       | c =  }} {{eqn | r = \dfrac {41} {42}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {41} {42}       | c =  }} {{eqn | r = \dfrac 1 {42}       | c =  }} {{end-eqn}} Thus we have: :$(1): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}$ $c = 8$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8       | c =  }} {{eqn | r = \dfrac {12 + 8 + 3} {24}       | c =  }} {{eqn | r = \dfrac {23} {24}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {23} {24}       | c =  }} {{eqn | r = \dfrac 1 {24}       | c =  }} {{end-eqn}} Thus we have: :$(2): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}$ $c = 9$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9       | c =  }} {{eqn | r = \dfrac {9 + 6 + 2} {18}       | c =  }} {{eqn | r = \dfrac {17} {18}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {17} {18}       | c =  }} {{eqn | r = \dfrac 1 {18}       | c =  }} {{end-eqn}} Thus we have: :$(3): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}$ $c = 10$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10}       | c =  }} {{eqn | r = \dfrac {15 + 10 + 3} {30}       | c =  }} {{eqn | r = \dfrac {28} {30}       | c =  }} {{eqn | r = \dfrac {14} {15}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {14} {15}       | c =  }} {{eqn | r = \dfrac 1 {15}       | c =  }} {{end-eqn}} Thus we have: :$(4): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}$ $c = 11$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {11}       | c =  }} {{eqn | r = \dfrac {33 + 22 + 6} {66}       | c =  }} {{eqn | r = \dfrac {61} {66}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {61} {66}       | c =  }} {{eqn | r = \dfrac 5 {66}       | c =  }} {{end-eqn}} But $\dfrac 5 {66}$ is not a unit fraction. Thus $a = 2, b = 3, c = 11$ does not lead to a solution. $c = 12$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12}       | c =  }} {{eqn | r = \dfrac {6 + 4 + 1} {12}       | c =  }} {{eqn | r = \dfrac {11} {12}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {11} {12}       | c =  }} {{eqn | r = \dfrac 1 {12}       | c =  }} {{end-eqn}} Thus we have: :$(5): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}$ Let $c > 12$. Then: :$1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c > \dfrac 1 {12}$ and so: :$d < c$ and so no further solutions can be found where $a = 2, b = 3, c > 12$. Hence there are exactly $5$ solutions where $a = 2, b = 3$. {{qed|lemma}} ==== $b = 4$ ==== Let $b = 4$. {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b       | r = \dfrac 1 2 + \dfrac 1 4       | c =  }} {{eqn | r = \dfrac {2 + 1} 4       | c =  }} {{eqn | r = \dfrac 3 4       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c + \dfrac 1 d       | r = 1 - \dfrac 3 4       | c =  }} {{eqn | r = \dfrac 1 4       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c       | o = <       | r = \dfrac 1 4       | c =  }} {{eqn | ll= \leadsto       | l = c       | o = >       | r = 4       | c =  }} {{end-eqn}} Thus we try $c = 5, 6, \ldots$ in turn. $c = 5$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5       | c =  }} {{eqn | r = \dfrac {10 + 5 + 4} {20}       | c =  }} {{eqn | r = \dfrac {19} {20}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {19} {20}       | c =  }} {{eqn | r = \dfrac 1 {20}       | c =  }} {{end-eqn}} Thus we have: :$(6): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}$ $c = 6$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6       | c =  }} {{eqn | r = \dfrac {12 + 6 + 4} {24}       | c =  }} {{eqn | r = \dfrac {22} {24}       | c =  }} {{eqn | r = \dfrac {11} {12}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {11} {12}       | c =  }} {{eqn | r = \dfrac 1 {12}       | c =  }} {{end-eqn}} Thus we have: :$(7): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}$ $c = 7$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 7       | c =  }} {{eqn | r = \dfrac {14 + 7 + 4} {28}       | c =  }} {{eqn | r = \dfrac {25} {28}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {25} {28}       | c =  }} {{eqn | r = \dfrac 3 {28}       | c =  }} {{end-eqn}} But $\dfrac 3 {28}$ is not a unit fraction. Thus $a = 2, b = 4, c = 7$ does not lead to a solution. $c = 8$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8       | c =  }} {{eqn | r = \dfrac {4 + 2 + 1} 8       | c =  }} {{eqn | r = \dfrac 7 8       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac 7 8       | c =  }} {{eqn | r = \dfrac 1 8       | c =  }} {{end-eqn}} Thus we have: :$(8): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8$ Let $c > 8$. Then: :$1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c > \dfrac 1 8$ and so: :$d < c$ and so no further solutions can be found where $a = 2, b = 4, c > 8$. Hence there are exactly $3$ solutions where $a = 2, b = 4$. {{qed|lemma}} ==== $b = 5$ ==== Let $b = 5$. {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b       | r = \dfrac 1 2 + \dfrac 1 5       | c =  }} {{eqn | r = \dfrac {5 + 2} {10}       | c =  }} {{eqn | r = \dfrac 7 {10}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c + \dfrac 1 d       | r = 1 - \dfrac 7 {10}       | c =  }} {{eqn | r = \dfrac 3 {10}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c       | o = <       | r = \dfrac 3 {10}       | c =  }} {{eqn | ll= \leadsto       | l = c       | o = \ge       | r = 5       | c =  }} {{end-eqn}} Thus we try $c = 5, 6, \ldots$ in turn. $c = 5$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5       | c =  }} {{eqn | r = \dfrac {7 + 2 + 2} {10}       | c =  }} {{eqn | r = \dfrac 9 {10}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac 9 {10}       | c =  }} {{eqn | r = \dfrac 1 {10}       | c =  }} {{end-eqn}} Thus we have: :$(9): \quad 1 = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}$ $c = 6$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 6       | c =  }} {{eqn | r = \dfrac {15 + 6 + 5} {30}       | c =  }} {{eqn | r = \dfrac {26} {30}       | c =  }} {{eqn | r = \dfrac {13} {15}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {13} {15}       | c =  }} {{eqn | r = \dfrac 2 {15}       | c =  }} {{end-eqn}} But $\dfrac 2 {15}$ is not a unit fraction. Thus $a = 2, b = 5, c = 6$ does not lead to a solution. $c = 7$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 7       | c =  }} {{eqn | r = \dfrac {35 + 14 + 10} {70}       | c =  }} {{eqn | r = \dfrac {59} {70}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {59} {70}       | c =  }} {{eqn | r = \dfrac {11} {70}       | c =  }} {{end-eqn}} But $\dfrac {11} {70}$ is not a unit fraction. Thus $a = 2, b = 5, c = 7$ does not lead to a solution. Let $c > 7$. {{begin-eqn}} {{eqn | l = 1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | o = >       | r = \dfrac {11} {70}       | c =  }} {{eqn | o = >       | r = \dfrac {10} {70}       | c =  }} {{eqn | r = \dfrac 1 7       | c =  }} {{end-eqn}} and so: :$d < c$ and so no further solutions can be found where $a = 2, b = 5, c > 7$. Hence there is exactly $1$ solution where $a = 2, b = 5$. {{qed|lemma}} ==== $b = 6$ ==== Let $b = 6$. {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b       | r = \dfrac 1 2 + \dfrac 1 6       | c =  }} {{eqn | r = \dfrac {3 + 1} 6       | c =  }} {{eqn | r = \dfrac 4 6       | c =  }} {{eqn | r = \dfrac 2 3       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c + \dfrac 1 d       | r = 1 - \dfrac 3 3       | c =  }} {{eqn | r = \dfrac 1 3       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c       | o = <       | r = \dfrac 1 3       | c =  }} {{eqn | ll= \leadsto       | l = c       | o = >       | r = 3       | c =  }} {{end-eqn}} But we already have that $c \ge b$. Thus: :$c \ge 6$ Thus we try $c = 6, 7, \ldots$ in turn. $c = 6$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6       | c =  }} {{eqn | r = \dfrac {3 + 1 + 1} 6       | c =  }} {{eqn | r = \dfrac 5 6       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac 5 6       | c =  }} {{eqn | r = \dfrac 1 6       | c =  }} {{end-eqn}} Thus we have: :$(10): \quad 1 = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6$ Let $c > 6$. Then: :$1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c > \dfrac 1 6$ and so: :$d < c$ and so no further solutions can be found where $a = 2, b = 6, c > 6$. Hence there is exactly $1$ solution where $a = 2, b = 6$. {{qed|lemma}} ==== $b > 6$ ==== Let $b > 6$. Then: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 d       | o = <       | r = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} Hence there are no solutions such that $a = 2, b > 6$ {{qed|lemma}} === $a = 3$ === Let $a = 3$. ==== $b = 3$ ==== Let $b = 3$. {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b       | r = \dfrac 1 3 + \dfrac 1 3       | c =  }} {{eqn | r = \dfrac 2 3       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c + \dfrac 1 d       | r = 1 - \dfrac 2 3       | c =  }} {{eqn | r = \dfrac 1 3       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c       | o = <       | r = \dfrac 1 3       | c =  }} {{eqn | ll= \leadsto       | l = c       | o = >       | r = 3       | c =  }} {{end-eqn}} Thus we try $c = 4, 5, \ldots$ in turn. $c = 4$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4       | c =  }} {{eqn | r = \dfrac {4 + 4 + 3} {12}       | c =  }} {{eqn | r = \dfrac {11} {12}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {11} {12}       | c =  }} {{eqn | r = \dfrac 1 {12}       | c =  }} {{end-eqn}} Thus we have: :$(11): \quad 1 = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}$ $c = 5$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 5       | c =  }} {{eqn | r = \dfrac {5 + 5 + 3} {15}       | c =  }} {{eqn | r = \dfrac {13} {15}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {13} {15}       | c =  }} {{eqn | r = \dfrac 2 {15}       | c =  }} {{end-eqn}} But $\dfrac 2 {15}$ is not a unit fraction. Thus $a = 3, b = 3, c = 5$ does not lead to a solution. $c = 6$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6       | c =  }} {{eqn | r = \dfrac {2 + 2 + 1} 6       | c =  }} {{eqn | r = \dfrac 5 6       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac 5 6       | c =  }} {{eqn | r = \dfrac 1 6       | c =  }} {{end-eqn}} Thus we have: :$(12): \quad 1 = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 6$ Let $c > 6$. Then: :$1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c > \dfrac 1 6$ and so: :$d < c$ and so no further solutions can be found where $a = 3, b = 3, c > 6$. Hence there are exactly $2$ solutions where $a = 3, b = 3$. {{qed|lemma}} ==== $b = 4$ ==== Let $b = 4$. {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b       | r = \dfrac 1 3 + \dfrac 1 4       | c =  }} {{eqn | r = \dfrac {4 + 3} {12}       | c =  }} {{eqn | r = \dfrac 7 {12}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c + \dfrac 1 d       | r = 1 - \dfrac 7 {12}       | c =  }} {{eqn | r = \dfrac 5 {12}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 c       | o = <       | r = \dfrac 5 {12}       | c =  }} {{eqn | ll= \leadsto       | l = c       | o = >       | r = \dfrac {12} 5       | c =  }} {{end-eqn}} But we already have that $c \ge b$. Thus: :$c \ge 4$ Thus we try $c = 4, 5, \ldots$ in turn. $c = 4$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4       | c =  }} {{eqn | r = \dfrac {4 + 3 + 3} {12}       | c =  }} {{eqn | r = \dfrac {10} {12}       | c =  }} {{eqn | r = \dfrac 5 6       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac 5 6       | c =  }} {{eqn | r = \dfrac 1 6       | c =  }} {{end-eqn}} Thus we have: :$(13): \quad 1 = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6$ $c = 5$: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | r = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5       | c =  }} {{eqn | r = \dfrac {20 + 15 + 12} {60}       | c =  }} {{eqn | r = \dfrac {47} {60}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 d       | r = 1 - \dfrac {47} {60}       | c =  }} {{eqn | r = \dfrac {13} {60}       | c =  }} {{end-eqn}} But $\dfrac {13} {60}$ is not a unit fraction. Thus $a = 3, b = 4, c = 5$ does not lead to a solution. Let $c > 5$. {{begin-eqn}} {{eqn | l = 1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c       | o = >       | r = \dfrac {13} {60}       | c =  }} {{eqn | o = >       | r = \dfrac {12} {60}       | c =  }} {{eqn | r = \dfrac 1 5       | c =  }} {{end-eqn}} and so: :$d < c$ and so no further solutions can be found where $a = 3, b = 4, 5 > 7$. Hence there is exactly $1$ solution where $a = 3, b = 4$. {{qed|lemma}} ==== $b > 4$ ==== Let $b > 4$. Then: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 d       | o = \le       | r = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 5       | c =  }} {{eqn | r = \dfrac {5 + 3 + 3 + 3} {15}       | c =  }} {{eqn | r = \dfrac {14} {15}       | c =  }} {{eqn | o = <       | r = 1       | c =  }} {{end-eqn}} Hence there are no solutions such that $a = 3, b > 5$. {{qed|lemma}} === $a = 4$ === Let $a = 4$. Let $d > 4$. Then: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 d       | o = <       | r = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4       | c =  }} {{eqn | o = <       | r = 1       | c =  }} {{end-eqn}} Thus the only solution where $a = 4$ is: :$(14): \quad 1 = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4$ {{qed|lemma}} === $a > 4$ === Suppose $a > 4$. Then: {{begin-eqn}} {{eqn | l = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 d       | o = <       | r = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4       | c =  }} {{eqn | o = <       | r = 1       | c =  }} {{end-eqn}} Hence there are no solutions such that $a > 4$. {{qed|lemma}} === Summary === Hence our $14$ solutions: {{begin-eqn}} {{eqn | n = 1       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}       | o =  }} {{eqn | n = 2       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}       | o =  }} {{eqn | n = 3       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}       | o =  }} {{eqn | n = 4       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}       | o =  }} {{eqn | n = 5       | r = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}       | o =  }} {{eqn | n = 6       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}       | o =  }} {{eqn | n = 7       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}       | o =  }} {{eqn | n = 8       | r = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8       | o =  }} {{eqn | n = 9       | r = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}       | o =  }} {{eqn | n = 10       | r = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6       | o =  }} {{eqn | n = 11       | r = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}       | o =  }} {{eqn | n = 12       | r = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}       | o =  }} {{eqn | n = 13       | r = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6       | o =  }} {{eqn | n = 14       | r = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4       | o =  }} {{end-eqn}} {{qed}}