Unit fractions with equal sum 1/a+1/b = 1/x+1/y?

Question

Characterize the quadruples (a,b,x,y) of integers 0<a<x≤y<b that satisfy the equation:

1/a+1/b = 1/x+1/y

The case a=2 has been completely covered in my previous question:
http://answers.yahoo.com/question/index;_ylt=AhZyhEpAT5zRyk_zHwl3He7sy6IX;_ylv=3?qid=20121118150240AAqJ3tz

But note how the number of solutions increases for a=3 as we have:

(3,b,4,y):
1/3+1/6 = 1/4+1/4
1/3+1/12 = 1/4+1/6
1/3+1/24 = 1/4+1/8
1/3+1/36 = 1/4+1/9
1/3+1/60 = 1/4+1/10
1/3+1/132 = 1/4+1/11

(3,b,5,y):
1/3+1/15 = 1/5+1/5
1/3+1/30 = 1/5+1/6
1/3+1/105 = 1/5+1/7

Is there any clever pattern to the solutions other than the fine restrictions given in the answer to my previous question? How can the method be extended to larger a's and how many solutions will there be for a given a>3?

Anonymous · Accepted Answer

Well, following az_lender's approach here is a partial answer... Suppose 0 < a < x <= y < b, and 1/a + 1/b = 1/x + 1/y. Then:

1/y - 1/b = 1/a - 1/x

1/y > 1/a - 1/x >= 1/a - 1/(a+1) = 1/[a(a+1)] ===> y < a(a+1)

1/x - 1/b = 1/a - 1/y

1/x > 1/a - 1/y >= 1/a - 1/(a+1) = 1/[a(a+1)] ====> x < a(a+1)

Thus, given a choice of a, we have:

a< x < a(a+1) [a^2 - 1 choices for x]

a < y < a(a+1) [a^2 - 1 choices for y]

Then b is determined once a,y,x are determined. So for a given a, there are at most (a^2-1)^2 solutions. Actually, a more accurate bound that considers the fact x <= y is sum_{x=a+1}^{a(a+1)-1} [a(a+1)-x] = (1/2)a^2(a^2-1).

Thus, there are at most (1/2)a^2(a^2-1) solutions for a given integer a. This allows a simple computer search. Writing the program gives the following:

***

a = 2: At most (1/2)a^2(a^2-1)=6 solutions.

1/2 + 1/6 = 1/3 + 1/3

1/2 + 1/12 = 1/3 + 1/4

1/2 + 1/30 = 1/3 + 1/5

Found 3 solutions

***

a = 3: At most (1/2)a^2(a^2-1)=36 solutions.

1/3 + 1/6 = 1/4 + 1/4

1/3 + 1/12 = 1/4 + 1/6

1/3 + 1/24 = 1/4 + 1/8

1/3 + 1/36 = 1/4 + 1/9

1/3 + 1/60 = 1/4 + 1/10

1/3 + 1/132 = 1/4 + 1/11

1/3 + 1/15 = 1/5 + 1/5

1/3 + 1/30 = 1/5 + 1/6

1/3 + 1/105 = 1/5 + 1/7

Found 9 solutions

***

a = 4: At most (1/2)a^2(a^2-1)=120 solutions.

1/4 + 1/20 = 1/5 + 1/10

1/4 + 1/30 = 1/5 + 1/12

1/4 + 1/60 = 1/5 + 1/15

1/4 + 1/80 = 1/5 + 1/16

1/4 + 1/180 = 1/5 + 1/18

1/4 + 1/380 = 1/5 + 1/19

1/4 + 1/12 = 1/6 + 1/6

1/4 + 1/24 = 1/6 + 1/8

1/4 + 1/36 = 1/6 + 1/9

1/4 + 1/60 = 1/6 + 1/10

1/4 + 1/132 = 1/6 + 1/11

1/4 + 1/28 = 1/7 + 1/7

1/4 + 1/56 = 1/7 + 1/8

1/4 + 1/252 = 1/7 + 1/9

Found 14 solutions

***

a = 5: At most (1/2)a^2(a^2-1)=300 solutions.

1/5 + 1/15 = 1/6 + 1/10

1/5 + 1/20 = 1/6 + 1/12

1/5 + 1/30 = 1/6 + 1/15

1/5 + 1/45 = 1/6 + 1/18

1/5 + 1/60 = 1/6 + 1/20

1/5 + 1/70 = 1/6 + 1/21

1/5 + 1/120 = 1/6 + 1/24

1/5 + 1/150 = 1/6 + 1/25

1/5 + 1/195 = 1/6 + 1/26

1/5 + 1/270 = 1/6 + 1/27

1/5 + 1/420 = 1/6 + 1/28

1/5 + 1/870 = 1/6 + 1/29

1/5 + 1/70 = 1/7 + 1/14

1/5 + 1/105 = 1/7 + 1/15

1/5 + 1/595 = 1/7 + 1/17

1/5 + 1/20 = 1/8 + 1/8

1/5 + 1/40 = 1/8 + 1/10

1/5 + 1/120 = 1/8 + 1/12

1/5 + 1/520 = 1/8 + 1/13

1/5 + 1/45 = 1/9 + 1/9

1/5 + 1/90 = 1/9 + 1/10

1/5 + 1/495 = 1/9 + 1/11

Found 22 solutions

***

For a=6, my program found 33 solutions.

For a=7, my program found 29 solutions.

For a=8, my program found 44 solutions.

For a=9, my program found 56 solutions.

For a=10, my program found 65 solutions.

...

For a=100, my program found 957 solutions, the first being 1/100 + 1/2525 = 1/101 + 1/2020 and the last being 1/100 + 1/3999900 = 1/199 + 1/201

***

Note: I am defining a,b,x,y as type "long" in C, and am using the condition x*y*b + a*x*y- a*b*y - a*b*x == 0 to test the values. The program seems to be working correctly (I spot checked a few of the solutions).

***

EDIT: For any solution (a,b,x,y), we have another solution (Ka, Kb, Kx, Ky) for any positive integer K. This may explain why the number of solutions takes a dip when a is a prime number greater than 5 (for example, for a=11 my program finds 50 solutions, for a=101 my program finds 175 solutions).

***

EDIT: Note that for any integer a>0 (including primes), the following is a solution:

1/a + 1/[a(2a-1)] = 1/(2a-1) + 1/(2a-1)

This shows that there are an infinite number of solutions that are not multiples of other solutions.

***

EDIT: If we fix a and assume b is a multiple of a, x, and y, then b = ka = wx = zy for some positive integers k, w, z. Then 1/a + 1/b = 1/x + 1/y is equivalent to k+1 = w + z. Then b = (w+z-1)a, and we have a solution if (w+z-1)a is divisible by both w and z. Equivalently, if we can find positive integers w, z such that the following two conditions hold:

(i) (w-1)a is divisible by z

(ii) (z-1)a is divisible by w

Then we have the solution 1/a + 1/[(w+z-1)a] = 1/[(w+z-1)a/w] + 1/[(w+z-1)a/z].

Using w=z=a gives the solution in the previous edit. Using z=a, w=a-1 gives the following solution:

1/a + 1/[(2a-2)a] = 1/(2a) + 1/(2a-2).

Using z=a, w=(a-1)a gives the following solution:

1/a + 1/[(a^2-1)a] = 1/(a+1) + 1/(a^2-1)

If we assume a>1 and fix z=a, then this procedure finds as many unique solutions as there are integers w>1 that divide (a-1)a. If a>2, there are always at least 3 solutions of this type (corresponding to w=a-1, w=a, w=a(a-1)). If a>3, there are always at least 4 solutions of this type (corresponding to w=2, w=a-1, w=a, w=a(a-1)).

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Unit fractions with equal sum 1/a+1/b = 1/x+1/y?

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