unit fraction

The Problem:
	A unit fraction is the reciprocal 1n of a positive integer n . The unit fraction 110 can be represented as a difference of unit fractions in the following four ways: 110=15−110;110=16−115;110=18−140;110=19−190. In how many ways can the fraction 12013 be expressed in the form 12013=1x−1y, where x and y are positive integers? Answer. In 13 ways.

The solution:

Because 1x=12013+1y, solving for x we get

x=2013yy+2013 (1)

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 20132 to both sides to get

20132==20132+2013y−x(y+2013)2013(2013+y)−x(2013+y)=(2013−x)(2013+y).

Because x and y are positive integers we see that 20132 has been written as the product of two positive integers, one strictly less than 2013 and one strictly greater. Moreover,

20132=32×112×612,

whence the number of factors of 20132 is (2+1)⋅(2+1)⋅(2+1)=27 , of which 13 exceed 2013 (and are candidates for 2013+y ), while 13 are less than 2013 (and serve
as 2013−x ). The thirteen possibilities are listed in the following table (taken from the solutions of Hovious and of Vause).

213−x	213+y	x	y	12013
1	4 052 169	2012	4 050 156	12012−14050156
3	1 350 723	2010	1 348 710	12010−11348710
32=9	450 241	2004	448 228	12004−1448228
11	368 379	2002	366 366	12002−1366366
3⋅11=33	122 793	1980	120 780	11980−1120780
61	66 429	1952	64 416	11952−164416
32⋅11=99	40 931	1914	38 918	11914−138918
112=121	33 489	1892	31 476	11892−131476
3⋅61=183	22 143	1830	20 130	11830−120130
3⋅112=363	11 163	1650	9150	11650−19150
32⋅61=549	7381	1464	5368	11464−15368
11⋅61=671	6039	1342	4026	11342−14026
32⋅112=1089	3721	924	1708	1924−11708

Method 2. We can simplify equation (1) a bit to get

x=2013yy+2013=2013−20132y+2013.

We deduce that x will be an integer if and only if the right-most fraction is an integer, which happens if and only if y+2013 divides evenly into 20132 . Moreover, because y is positive we have y+2013>2013 , which for one thing forces x to be positive (because 20132y+2013<2013 ); another consequence is that each factor f(=y+2013) of 20132 that exceeds 2013 will produce a pair of unit fractions that satisfy all our conditions, namely 1x=(2013−20132f)−1 and (because y=f−2013 ) 1y=1f−2013 .



Comments. Problem 2175 in the problem-solving journal Crux Mathematicorum with Mathematical Mayhem, 23:7 (November 1997), pages 443-444, 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 1n can be written as a difference of two unit fractions in τ(n2)−12 ways, where τ(n2) is the number of divisors of n2 . Specifically, if n2=p2e11⋅p2e22⋅…⋅p2ekk is the prime factorization of n2 , then τ(n2)=(2e1+1)(2e2+1)…(2ek+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.

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