An Equation in Reciprocals

Prove that for all integer n ≥ 3, the equation

1/x₁ + 1/x₂ + ... + 1/x_n = 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/x_n = 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/x_n = 1. Then 1/2x₁ + 1/2x₂ + ... + 1/2x_n = 1/2 and, therefore,

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

making (2, 2x₁, 2x₂, ..., 2x_n) a solution for n + 1, because, obviously all the terms (2, 2x₁, 2x₂, ..., 2x_n) 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, ..., 2^{n - 2}, 2^{n - 2} + 1, 2^{n - 2}(2^{n - 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/(2^n-2 + 1) + 1/2^{n - 2}(2^{n - 2} + 1)) = 1/(2^n-2 + 1) + 1/2^{n - 2} - 1/(2^{n - 2} + 1) = 1/2^{n - 2}.

Adding up the latter two proves the claim.

Solution 4

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

a_k+1 - 1 = a_k(a_k - 1), k ≥ 1.

In other words,

1 / a_k = 1 / (a_k - 1) - 1 / (a_{k + 1} - 1).

So that the sum 1/a₁ + ... + 1/a_{n - 1} telescopes:

1/a₁ + ... + 1/a_{n - 1} = 1 / (a₁ - 1) - 1 / (a_{n - 1} - 1) = 1 - 1 / (a_{n - 1} - 1).

Thus, for n ≥ 3, we get solutions (a₁ = 2, a₂, ..., a_{n - 1}, a_n - 1).

Solution 5

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

(a₁, a₂, ..., a_n-1, a_n + 1, a_n(a_n + 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 / a^m + 1 / a^m(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, ..., 7^n-3, 6·7^n-2).

References

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

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