## An Equation in Reciprocals

Prove that for all integer n ≥ 3, the equation

1/x1 + 1/x2 + ... + 1/xn = 1

is solvable in distinct positive integers.

Example: (2, 3, 6) solve the equation for n = 3: 1/2 + 1/3 + 1/6 = 1.

Solution

Prove that for all integer n ≥ 3, the equation

1/x1 + 1/x2 + ... + 1/xn = 1

is solvable in distinct positive integers.

### Solution 1

Here is a proof by induction. Assume, for a given n ≥ 3, 1/x1 + 1/x2 + ... + 1/xn = 1. Then 1/2x1 + 1/2x2 + ... + 1/2xn = 1/2 and, therefore,

1/2 + 1/2x1 + 1/2x2 + ... + 1/2xn = 1

making (2, 2x1, 2x2, ..., 2xn) a solution for n + 1, because, obviously all the terms (2, 2x1, 2x2, ..., 2xn) are distinct.

### Solution 2

The sum below is from k = 1 through k = n-1:

∑k / (k + 1)! = ∑(k + 1 - 1) / (k + 1)! = ∑(1 / k! - 1 / (k + 1)!) = 1 - 1/n!.

From here (n!, 2!/1, 3!/2, ..., (k+1)!/k, ..., n!/(n-1)) solve the equation for n.

### Solution 3

Another solution is given by (2, 22, ..., 2n - 2, 2n - 2 + 1, 2n - 2(2n - 2 + 1)).

To see why is it so, first evaluate

2-1 + 2-2 + ... + 2-(n-2) = 1/2 · (1 - 2-(n-2))/(1 - 1/2) = 1 - 2-(n-2).

Next, consider

1/(2n-2 + 1) + 1/2n - 2(2n - 2 + 1)) = 1/(2n-2 + 1) + 1/2n - 2 - 1/(2n - 2 + 1) = 1/2n - 2.

Adding up the latter two proves the claim.

### Solution 4

From a sequence a1 = 2, am+1 = a1·...·am + 1, m ≥ 1. It follows from the recurrence that

ak+1 - 1 = ak(ak - 1), k ≥ 1.

In other words,

1 / ak = 1 / (ak - 1) - 1 / (ak + 1 - 1).

So that the sum 1/a1 + ... + 1/an - 1 telescopes:

1/a1 + ... + 1/an - 1 = 1 / (a1 - 1) - 1 / (an - 1 - 1) = 1 - 1 / (an - 1 - 1).

Thus, for n ≥ 3, we get solutions (a1 = 2, a2, ..., an - 1, an - 1).

### Solution 5

If (a1, a2, ..., an-1, an) is a solution for a given n, with a1 < a2 < ... < an, then

(a1, a2, ..., an-1, an + 1, an(an + 1))

is a solution for n + 1.

### Solution 6

The most common telescoping series also leads to a solution (2, 6, 12, ..., (n - 1)n, n(n + 1), (n + 1)).

### Solution 7

For a > 1, the following formula

1 / (a - 1) = 1 / a + 1 / a² + ... + 1 / am + 1 / am(a - 1)

may be used to generate more solutions from the one given. The trick to replace the fraction with the largest denominator using the above formula. For example, with a = 7, solution (2, 3, 6) leads to (2, 3, 7, 72, ..., 7n-3, 6·7n-2).

### References

1. T. Andreescu, D. Andrica, I. Cucurezeanu, An Introduction to Diophantine Equations, Birkhäuser, 2010, pp. 38-42

• Diophantine Equations
• Finicky Diophantine Equations
• Diophantine Quadratic Equation in Three Variables
• A Short Equation in Reciprocals
• Minus One But What a Difference
• Two-Parameter Solutions to Three Almost Fermat Equations
• Chinese Remainder Theorem
• Step into the Elliptic Realm
• Fermat's Like Equation
• Sylvester's Problem
• Sylvester's Problem, A Second Look
• Negative Coconuts