## An Equation in Reciprocals

Prove that for all integer n ≥ 3, the equation

1/x_{1} + 1/x_{2} + ... + 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.

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Copyright © 1996-2018 Alexander BogomolnyProve that for all integer n ≥ 3, the equation

1/x_{1} + 1/x_{2} + ... + 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} + 1/x_{2} + ... + 1/x_{n} = 1._{1} + 1/2x_{2} + ... + 1/2x_{n} = 1/2

1/2 + 1/2x_{1} + 1/2x_{2} + ... + 1/2x_{n} = 1

making (2, 2x_{1}, 2x_{2}, ..., 2x_{n}) a solution for _{1}, 2x_{2}, ..., 2x_{n})

### 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, 2^{2}, ..., 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 a_{1} = 2, a_{m+1} = a_{1}·...·a_{m} + 1,

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/a_{1} + ... + 1/a_{n - 1} telescopes:

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

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

### Solution 5

If (a_{1}, a_{2}, ..., a_{n-1}, a_{n}) is a solution for a given n, with _{1} < a_{2} < ... < a_{n},

(a_{1}, a_{2}, ..., 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

### 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}, ..., 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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Copyright © 1996-2018 Alexander Bogomolny63419720 |