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.
|Contact| |Front page| |Contents| |Arithmetic| |Up|
Copyright © 1996-2018 Alexander BogomolnyProve 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,
1/2 + 1/2x1 + 1/2x2 + ... + 1/2xn = 1
making (2, 2x1, 2x2, ..., 2xn) a solution for
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,
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-1, an + 1, an(an + 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 / 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
References
- T. Andreescu, D. Andrica, I. Cucurezeanu, An Introduction to Diophantine Equations, Birkhäuser, 2010, pp. 38-42
|Contact| |Front page| |Contents| |Arithmetic| |Up|
Copyright © 1996-2018 Alexander Bogomolny71537157