Talk:Erdős–Straus conjecture

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Geometry guy 21:48, 9 July 2007 (UTC)

The Swett link appears to be broken.

[edit] Correctness of a related conjecture

I very much doubt the correctness of this section in Generalization: Hagedorn proved a related conjecture of R. H. Hardin and Neil Sloane that, for odd positive n, the equation 3/n = 1/x + 1/y + 1/z is always solvable with x, y, and z also odd and positive. A proof of this conjecture would be trivial: Let $x = y = z$ be $n$ . Ocolon 22:33, 29 January 2007 (UTC)

x, y, and z must not equal each other. I edited in a clarification. —David Eppstein 03:16, 30 January 2007 (UTC)

[edit] Link to Leibniz harmonic triangle

A link was added to the Leibniz harmonic triangle. I wonder if this is relevant to the topic. The Leibniz harmonic triangle might be used to find solutions for $n = 4 m$ but the case is rather simple anyway: $\frac{4}{n}=\frac{1}{n}+\frac{1}{n}+\frac{1}{\frac{n}{2}}$ Ocolon 08:28, 4 March 2007 (UTC)

You stopped with that one case, but it can also be used to get solutions with distinct denominators. And with a little creativity, it can also help when 4|n is false. Anton Mravcek 20:10, 4 March 2007 (UTC)

While this might be true, neither the description of the link nor the article about the Leibniz harmonic triangle says that it can be used to find solutions for natural numbers that are not divisible by 4 or that it provides distinct denominators. Besides, the Erdős–Straus conjecture does not require distinct denominators. Furthermore, I don't understand this article as collection of solutions for special cases of the Erdős–Straus conjecture. Shall we add links that solve it for all Mersenne primes, for all numbers that are divisible by 1234, for all square numbers…

I don't question that the Leibniz harmonic triangle may be used to compute some solutions. I question the relevance of that — and even if it was relevant — the helpfulness of the link as the Leibniz harmonic triangle article stops where I stopped: Naming that simple case. Ocolon 08:32, 5 March 2007 (UTC)

I was wondering the same thing. Also, are there published sources connecting the Leibniz triangle to the 4/n problem, or is it just someone's original research? —David Eppstein 18:28, 4 March 2007 (UTC)

There probably aren't any published sources making the connection. It's too simple. The first Google Scholar result uses a complicated bunch of equations to show why the triangle is symmetrical. Even so, I wouldn't be as dismissive as calling this "just someone's original research." It's probably been rediscovered several times by several different people. Anton Mravcek 20:10, 4 March 2007 (UTC)

[edit] A073101 does require that x < y < z

The sequence A073101 requires that $x < y < z$ . In particular, this means that $x$ , $y$ and $z$ must be distinct, something which is not required by the Erdös-Straus formulation. --Kuifware (talk) 19:02, 28 March 2008 (UTC)

If there exists a solution 4/n = 1/x + 1/y + 1/z with x,y,z non-distinct, then there exists another solution with them distinct. See e.g. [1]. —David Eppstein (talk) 19:17, 28 March 2008 (UTC)

You are right: for $n \ge 3$ you can construct a solution with distinct denominators from a solution where they are not necessarily distinct. (For

n = 2

this is not possible however.)

There is a difference with the proof at your website: for the

4 / n

problem the Egyptian fraction should always consist of exactly 3 terms, so the reduction from

1 / y + 1 / y

2 / y

for even

y

needs to be modified a bit. For example: write

y = 2 p y'

with the integer $p\ge0$ as large as possible (i.e.

y'

is odd), then the reduction $\textstyle \frac{1}{y} + \frac{1}{y} = \frac{1}{2^p} \cdot \frac{2}{y'} = \frac{2}{2^p(y'+1)} + \frac{2}{2^py'(y'+1)}$ can be used if

y' > 1

, and the reduction $\textstyle \frac{1}{y} + \frac{1}{y} = \frac{1}{2} \cdot \frac{1}{2^{p-2}} = \left(\frac{1}{3} + \frac{1}{6}\right) \cdot \frac{1}{2^{p-2}} = \frac{1}{3\cdot2^{p-2}} + \frac{1}{6\cdot2^{p-2}}$ can be used if

y' = 1

and $p\ge2$ . The remaining two cases, where $y \in \{1,2\}$ , are not covered because there are no Egyptian fractions representations for the numbers 1 and 2 using only two terms.

I found the argument for the finiteness of the reduction procedure interesting, though perhaps not entirely obvious. But I agree: there can only be a finite number of steps because the steps are lexicographically decreasing (the implication can be proven by induction on the number of terms). --85.144.141.41 (talk) 18:21, 29 March 2008 (UTC)

I slightly modified the text to reflect the fact that A073101 considers distinct denominators. I did not add a statement pointing out the subtle difference. --Kuifware (talk) 19:13, 29 March 2008 (UTC)

The change looks fine to me. —David Eppstein (talk) 19:46, 29 March 2008 (UTC)

Talk:Erdős–Straus conjecture

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[edit] Correctness of a related conjecture

[edit] Link to Leibniz harmonic triangle

[edit] A073101 does require that x < y < z

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