Condition for 3 over n producing 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient

Theorem
Consider proper fractions of the form $\dfrac 3 n$ expressed in canonical form.

Let Fibonacci's Greedy Algorithm be used to generate a sequence $S$ of Egyptian fractions for $\dfrac 3 n$.

Then $S$ consists of $3$ terms, where $2$ would be sufficient the following conditions hold:


 * $n \equiv 1 \pmod 6$
 * $\exists d: d \divides n$ and $d \equiv 2 \pmod 3$

Corollary
The smallest $n$ for which $S$ consists of $3$ terms, where $2$ would be sufficient, is $25$.

Proof
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 $n = 3 k - 2$.

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 $n$ is odd.

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:

Since $\dfrac 3 n$ is in canonical form, $x y \divides n$.

By Divisors of Product of Coprime Integers, we can find $p, q \in \N$ such that:
 * $p \divides x$, $q \divides y$
 * $p q = n$

Rewrite:

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:
 * $\dfrac 3 n = \dfrac 3 {n \paren {d + 1} } + \dfrac {3 d} {n \paren {d + 1} }$

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:

which is a contradiction.

Therefore $n$ cannot be expressed as the sum of $2$ Egyptian fractions.

Hence the result.