TITLE: An odd property of Egyptian fractions
QUESTION [8 upvotes]: This question arose through a response to this post.
For which integers $N>1$ does the fraction $\frac 1N$ appear in the Egyptian Fraction expansion of $\frac {N-1}{N}$?
To specify: As such expansions are not unique, I should say which one I refer to. Here we consider the expansion obtained through the greedy algorithm.  
Thus $$\frac 12=\frac 12\;\;\&\;\;\frac 34=\frac 12+\frac 14\;\;\&\;\;\frac {11}{12}=\frac 12+\frac 13+\frac 1{12}$$ are easy examples.
A quick search for $N<100$ yields $N=\{2,4,12,84\}$ as examples.  Taking that (short) list to OEIS leads to $[A053631][1]$, the sequence $a_i$ starting with $a_1=2$ and having the property that, for $i>1$, $\{a_{i-1}+1,a_i,a_i+1\}$ are a Pythagorean triple. That sequence continues from $84$ as $3612,\, 6526884,\, 21300113901612,\dots$ and it is easy to verify that those three, at least, are examples for the present question as well.
Are these all examples?  Are there others?
Edit:  as remarked in the comments, in each of the cases cited above, $\frac 1N$ appears as the final term in the expansion.

REPLY [4 votes]: Note that $2 \cdot 2=4, 4 \cdot 3=12, 12 \cdot 7=84, 84 \cdot 43=3612, 3612 \cdot 1807=6526884$.  If $k$ is a solution, then it is $\frac {k-2}k + \frac 1k$, where we don't care how $\frac {k-2}k$ is expressed.  Then let $m=k(\frac k2+1)$, which is a multiple of $4$ because $k$ is.  We have $\frac {m-1}m=\frac {k(\frac k2+1)-1}{k(\frac k2+1)}=\frac {k-2}k+\frac 1k+\frac 1{k+2}=\frac {k-2}k+\frac 1{k/2+1}+\frac 1{k(\frac k2+1)}=\frac {k-2}k+\frac 1{k/2+1}+\frac 1m$, so $m$ is a solution.  This shows there are an infinite number of solutions, but does not show there are no others