On a Problem About Odd-denominator Egyptian Fractions y 1
Abstract
The article provides a solution to a Diophantine equation, where the sum of the reciprocal values of n distinct odd positive integers is equal to 1.
The problem we solve is mentioned in Titu Andreescu’s & Dorin Andrica’s book (see [1]) and is a modification of the Egyptian Fractions problem, asking only for odd integers in the denominators of the decomposition.
Problem.
Determine all integers n > 1 for which the equation
1 x
1
+
1 x
2
1
+ · · · + x n
= 1 is solvable in distinct odd positive integers.
Proof.
We may assume x
1
< x
2
< . . . < x n
. First, we note that n is odd, since x
1 x
2
. . . x n
= x
2
. . . x n
+ · · · + x
1 x
2
. . . x n − 1 and the right hand-side is a sum of n odd numbers, while the left hand-side is an odd number.
If 1 < n ≤ 6, then x
1 must be larger than 1, but for the next six odd numbers we have
6
X
1 k =1
2 k + 1
< 1 .
Hence we have no solutions in this case.
If n = 7, then x
7 must be the largest number. Selecting x
1 to larger than 1 and pairwise distinct gives the following estimate x
5 the lowest possible odd integers
1 x
6
1
+ x
7
= 1 −
1
3
+
1
5
+
1
7
+
1
9
1
+
11
It follows that x
6
< 17. But for x
6
= 13 or 15 we have x
1
7
= 1 −
1
3
+
1
5
+
1
7
+
1
9
1
+
11
1
+
13
>
2
17
.
=
2021
45045
,
1 x
7
= 1 −
1
3
+
1
5
+
1
7
+
1
9
1
+
11
1
+
15
=
191
3465
, both equalities being impossible. But if x
1 to x
5 are not the lowest possible numbers, then
1 x
6
1
+ x
7
≥ 1 −
1
3
+
1
5
+
1
7
+
1
9
+
1
13
2
>
15 and so x
6
< 15. But x
5
≥ 13, so x
5
≥ x
6 and that is not possible. Therefore n = 7 is not a solution.
1
Mathematical Reflections
5 (2013)

We now prove by induction on n that, if n ≥ 9 is odd, then we can find a solution of the equation for which 3 divides x n
.
If n = 9, consider the equality
1
3
+
1
5
+
1
7
+
1
9
1
+
11
1
+
15
1
+
21
1
+
165
1
+
693
= 1 .
Note that 3 divides 693. If x
1
, . . . , x n for i < n and y n
= is a solution to the equation with 3
5
3 x n
, y n +1
= 3 x n
, y n +2
= 15 x n
| x n
, then setting y i
= x i yields a solution of the equation
1 y
1
1
+ y
2
1
+ · · · + y n +2
= 1 , where 3 | y n +2
. This concludes the inductive step and shows that the solutions of the problem are the odd numbers greater than or equal to 9.
Acknowledgements thank Michal Rol´ınek for helping me to polish this article.
References
[1] T. Andreescu and D. Andrica, An Introduction to Diophantine Equations , GIL publishing house,
2002.
Mathematical Reflections
5 (2013)