Solution for March 2013
The Problem: 

A unit fraction is the reciprocal $\frac1n$ of a positive integer $n$. The unit fraction $\frac1{10}$ can be represented as a difference of unit fractions in the following four ways:
$$\frac1{10}=\frac15\frac1{10}; \quad \frac1{10}=\frac16\frac1{15}; \quad
\frac1{10}=\frac18\frac1{40}; \quad \frac1{10}=\frac19\frac1{90}. $$
In how many ways can the fraction $\frac1{2013}$ be expressed in the form
$$\frac1{2013} = \frac1x  \frac1y ,$$
where $x$ and $y$ are positive integers?
Answer. In 13 ways.
Lamis Alsheikh
(Syria) 
Jose Arraiz
(France) 
Ricardo Bernabé Baloni
(Argentina) 
Luigi Bernardini
(Italy) 
Aleksandar Blazhevski
(Macedonia) 
Radouan Boukharfane
(Québec) 
Lou Cairoli
(USA) 
Saturnino Campo Ruiz
(Spain) 
Ioan Viorel Codreanu
(Romania) 
Ruben Victor Cohen
(Argentina) 
Bernard Collignon
(France) 
Olivier Cyr
(France) 
Hubert Desprez
(France) 
Jean Drabbe
(Belgium) 
Allen Druze
(USA) 
MeiHui Fang
(Austria) 
Federico Foieri
(Argentina) 
Philippe Fondanaiche
(France) 
Jan Fricke
(Germany) 
Georges Ghosn
(Québec) 
Pierre Gobin
(France) 
Gruian Cornel
(Romania) 
Tony Harrison
(UK) 
James W. Hovious
(USA) 
Gilbert Julia
(France) 
Farid Lian
(Columbia) 
Matthew Lim
(USA) 
Nawal Kishor Mishra
(India) 
Muralidharan Prasaanth (USA) 
Sanjeev Ramachandra Nimishakavi 
Nicolás Otero 
Don Redmond
(USA) 
Mathias Schenker
(Switzerland) 
Heri Setiyawan
(Indonesia) 
Albert Stadler
(Switzerland) 
Hakan Summakoğlu
(Turkey) 
Bruno Tisserand
(France) 
Authur Vause
(UK) 

We also received four solutions that were incomplete and one that was incorrect.

The solution:
Because $\frac1x = \frac1{2013} + \frac1y ,$ solving for $x$ we get
\begin{equation}
\hspace{25 mm} x = \frac{2013y}{y+2013} \hspace{30 mm} \mbox{ (1)}
\end{equation}
From here the arguments from our correspondents took one of two paths.
Method 1. Clearing the denominator, we see that equation (1) becomes $0=2013y  x(y+2013)$. Now add $2013^2$ to both sides to get
\begin{eqnarray*}
2013^2 &=& 2013^2 + 2013yx(y+2013)\\
&=& 2013(2013+y)  x(2013+y) = (2013x)(2013+y).
\end{eqnarray*}
Because $x$ and $y$ are positive integers we see that $2013^2$ has been written as the product of two positive integers, one strictly less than 2013 and one strictly greater. Moreover,
$$2013^2 = 3^2 \times 11^2 \times 61^2,$$
whence the number of factors of $2013^2$ is $(2+1)\cdot(2+1)\cdot(2+1) = 27$, of which 13 exceed 2013 (and are candidates for $2013+y$), while 13 are less than $2013$ (and serve
as $2013x$). The thirteen possibilities are listed in the following table (taken from the solutions of Hovious and of Vause).
$\mathbf{213x}$ 
$\mathbf{213+y}$ 
$\mathbf{x}$ 
$\mathbf{y}$ 
$\mathbf{\frac1{2013}}$ 
1 
4 052 169 
2012 
4 050 156 
$\frac1{2012}  \frac1{4050156}$ 
3 
1 350 723 
2010 
1 348 710 
$\frac1{2010}  \frac1{1 348 710}$ 
$3^2 = 9$ 
450 241 
2004 
448 228 
$\frac1{2004}  \frac1{448 228}$ 
11 
368 379 
2002 
366 366 
$\frac1{2002}  \frac1{366 366}$ 
$ 3 \cdot 11 = 33$ 
122 793 
1980 
120 780 
$\frac1{1980}  \frac1{120 780}$ 
61 
66 429 
1952 
64 416 
$\frac1{1952}  \frac1{64 416}$ 
$3^2\cdot 11 = 99$ 
40 931 
1914 
38 918 
$\frac1{1914}  \frac1{38 918}$ 
$11^2 = 121$ 
33 489 
1892 
31 476 
$\frac1{1892}  \frac1{31 476}$ 
$3 \cdot 61 = 183$ 
22 143 
1830 
20 130 
$\frac1{1830}  \frac1{20 130}$ 
$3 \cdot 11^2 = 363$ 
11 163 
1650 
9150 
$\frac1{1650}  \frac1{9150}$ 
$3^2 \cdot 61 = 549$ 
7381 
1464 
5368 
$\frac1{1464}  \frac1{5368}$ 
$11 \cdot 61 = 671$ 
6039 
1342 
4026 
$\frac1{1342}  \frac1{4026}$ 
$3^2 \cdot 11^2 = 1089$ 
3721 
924 
1708 
$\frac1{924}  \frac1{1708}$ 
Method 2. We can simplify equation (1) a bit to get
$$x = \frac{2013y}{y+2013} = 2013  \frac{2013^2}{y+2013}.$$
We deduce that $x$ will be an integer if and only if the rightmost fraction is an integer, which happens if and only if $y+2013$ divides evenly into $2013^2$. Moreover, because $y$ is positive we have $y+2013 > 2013$, which for one thing forces $x$ to be positive (because $\frac{2013^2}{y+2013} < 2013$); another consequence is that each factor $f \; \; (= y+2013)$ of $2013^2$ that exceeds $2013$ will produce a pair of unit fractions that satisfy all our conditions, namely $\frac1x = \left(2013  \frac{2013^2}f\right)^{1}$ and (because $y = f2013$) $\frac1y = \frac1{f2013}$.
Comments. Problem 2175 in the problemsolving journal Crux Mathematicorum with Mathematical Mayhem, 23:7 (November 1997), pages 443444, is the source of our March problem; it was proposed by Christopher J. Bradley. The featured solution there came from Kipp Johnson, who is a regular contributor to our monthly solutions! His solution (which both then and now is quite similar to our Method 2), as well as many of our submitted solutions, showed more generally how any unit fraction $\frac1n$ can be written as a difference of two unit fractions in $\frac{\tau(n^2) 1}{2}$ ways, where $\tau(n^2)$ is the number of divisors of $n^2$. Specifically, if $n^2=p_1^{2e_1}\cdot p_2^{2e_2}\cdot\dots\cdot p_k^{2e_k}$ is the prime factorization of $n^2$, then $\tau(n^2) = (2e_1+1)(2e_2+1)\dots(2e_k+1)$. By coincidence, at around the same time the same problem appeared in the American Mathematical Monthly, 105:4 (April 1998) p. 372; the solution published there was much like our Method 1. Actually, the problem is much older: Drabbe informed us of an 1896 appearance in Dujardin, L'Intermédiaire des mathématiciens, tome III, page 14. However, we know that the ancient Egyptians worked with unit fractions, so it is quite possible that the problem was first posed and solved a couple thousand years ago.
