By Upper Limit of Number of Unit Fractions to express Proper Fraction from Greedy Algorithm, $S$ consists of no more than $3$ terms. Suppose $n$ has our desired property. Since $\dfrac 3 n$ is proper, $n \ge 4$. Since $\dfrac 3 n$ is in canonical form, $3 \nmid n$. We also have that $S$ consists of at least $2$ terms. Consider the case $n = 3 k - 1$. :$\dfrac 3 n = \dfrac 1 k + \dfrac 1 {k \paren {3 k - 1} } \quad$ as $\ceiling {\dfrac n 3} = \ceiling {k - \dfrac 1 3} = k$ Fibonacci's Greedy Algorithm produces $2$ terms only. hence it must be the case that $n = 3 k - 2$. {{qed|lemma}} We have: :$\dfrac 3 n = \dfrac 1 k + \dfrac 2 {k \paren {3 k - 2} } \quad$ as $\ceiling {\dfrac n 3} = \ceiling {k - \dfrac 2 3} = k$ If $k$ is even, $\dfrac 1 {\paren {k / 2} \paren {3 k - 2} }$ is an Egyptian fraction. Then Fibonacci's Greedy Algorithm would produce $2$ terms only. hence it must be the case that $k$ is odd. This happens {{iff}} $n$ is odd. {{qed|lemma}} We have shown that for Fibonacci's Greedy Algorithm to produce $3$ terms, $n$ must be odd and $n = 3 k - 2$. By Chinese Remainder Theorem, these conditions can be merged into: :$n \equiv 1 \pmod 6$ We need find when Fibonacci's Greedy Algorithm gives minimum terms. Write: {{begin-eqn}} {{eqn | l = \frac 3 n       | r = \frac 1 x + \frac 1 y }} {{eqn | r = \frac {x + y} {x y} }} {{end-eqn}} Since $\dfrac 3 n$ is in canonical form, $x y \divides n$. By Divisor of Product, we can find $p, q \in \N$ such that: :$p \divides x$, $q \divides y$ :$p q = n$ Rewrite: {{begin-eqn}} {{eqn | l = \frac 3 n       | r = \frac 1 {p a} + \frac 1 {q b} }} {{eqn | r = \frac {q b + p a} {p q a b} }} {{eqn | ll= \leadsto       | l = p a + q b       | r = 3 a b }} {{eqn | ll= \leadsto       | l = p       | r = \frac {b \paren {3 a - q} } a }} {{end-eqn}} If $\exists d: d \divides n$ and $d \equiv 2 \pmod 3$, set $q = 1$ and $a = \dfrac {d + 1} 3$. Then $p = n$ and $b = \dfrac {a n} {3 a - 1} = \dfrac {n \paren {d + 1} } {3 d}$, which is a solution: :$\dfrac 3 n = \dfrac 3 {n \paren {d + 1} } + \dfrac {3 d} {n \paren {d + 1} }$ {{qed|lemma}} Now suppose $d$ does not exist. Then any divisor of $n$ is equivalent to $1 \pmod 3$. Hence $q \equiv 1 \pmod 3$. Thus $3 a - q \equiv 2 \pmod 3$. Let $r = \gcd \set {3 a - q, a}$. We have: :$a \divides b \paren {3 a - q} = b r \paren {\dfrac {3 a - q} r}$ Thus: :$\dfrac a r \divides b \paren {\dfrac {3 a - q} r}$ By Integers Divided by GCD are Coprime: :$\dfrac a r \perp \dfrac {3 a - q} r$ Finally, by Euclid's Lemma: :$\dfrac a r \divides b$ Hence $\dfrac {3 a - q} r, \dfrac {b r} a \in \Z$, and $p = \paren {\dfrac {3 a - q} r} \paren {\dfrac {b r} a}$. Thus we also have: :$\dfrac {3 a - q} r \divides p \divides n$ :$r = \gcd \set {q, a} \divides q \divides n$ Hence: :$\dfrac {3 a - q} r \equiv 1 \pmod 3$ :$r \equiv 1 \pmod 3$ Taking their product: {{begin-eqn}} {{eqn | l = 3 a - q       | r = \frac {3 a - q} r \times r }} {{eqn | o = \equiv       | r = 1 \times 1       | rr = \pmod 3 }} {{eqn | o = \equiv       | r = 1       | rr = \pmod 3 }} {{end-eqn}} which is a contradiction. Therefore $n$ cannot be expressed as the sum of $2$ Egyptian fractions. Hence the result. {{qed}} Category:Fibonacci's Greedy Algorithm Category:Egyptian Fractions lkbiny67byhfdimpr3o1qt6xwno8ec4