# Difficulty with the Erdos-Straus Conjecture

1. Oct 4, 2003

### Descender

The Erdos-Straus Conjecture proposes that:

For n >= 1, 4/n = 1/a + 1/b + 1/c, has postive integer solutions.
n = 2k (evens)
n = 4k+3 (odds)
n = 4k+1 (odds)

For 2k I have found the pattern --> if k >= 1, 4/2k = 1/2k + 1/2k + 1/k. This is simple, I know.

I am having difficulty finding the pattern for 4k+3 (no pattern has yet been found for 4k+1, I know).

So what is the pattern for:

4/4k+3 = 1/a + 1/b + 1/c

?

Does anyone know this pattern? If you do, please reply with such pattern.

Thank you
J.E.H.

2. Oct 5, 2003

### MathematicalPhysicist

n>=1
n=2k
it should be 2k>=1 not k>=1

3. Oct 5, 2003

### MathematicalPhysicist

maybe: 4/4k+3=1/4k+3+1/4k+3+1/0.5*(4k+3)

but im not sure this what you are after.

4. Nov 19, 2004

### jpease

The denominator of the last term in loop quantum gravity's equation [0.5*(4k+3)] will never be an integer, so it's not a valid unit fraction.

The {a,b,c} triplet I came up with for 4/(4k+3) is:

{ 2(k+1), 2(k+1), (k+1)(4k+3) }

This works too: { k+1, 2(k+1)(4k+3), 2(k+1)(4k+3) }.

These two triplets have the same source: Substract 1/(k+1) from 4/(4k+3) and you get 1/(k+1)(4k+3). But since you now have an equation of the form

4/(4k+3) = 1/a + 1/b

you need to add a third term somehow to "fit" the conjecture. You can do so by splitting one of the right-hand terms as follows:

1/a = 1/(2a) + 1/(2a).

I've worked on this conjecture off and on the last 20 years but have never proved it. But I have found many results, including that the smallest {n} that is a counterexample--if one exists--is prime and equal to 1 mod 24.

Last edited: Nov 19, 2004
5. Nov 19, 2004

### vsage

What a weird conjecture. Anyway 4k+1 with trivial solutions (making the substitution k = k - 0.5) is $$\frac {4}{4k+1} = \frac {1}{2(k+0.5)(4k+1)} + \frac {1}{2(k+0.5)(4k+1)} + \frac {1}{k+0.5}$$. However, for n mod 4 = 1 (5, for example) neither this equation or the author above's works. Allan Swett (http://math.uindy.edu/swett/esc.htm) appears to be the most learned in this subject. Perhaps his page will give you some insights. Also, your solution to 1/2k I don't think is entirely correct because egyptian fractions are unique usually.

Last edited by a moderator: Nov 19, 2004
6. Aug 22, 2005

### bruno C

It's not that hard to find three different Egyptian fractions that add up to
4/(4k+3).

For any fraction of the form 4/(4k+3), compute the difference with the nearest integer fraction:

4/(4k+3) - 1/(k+1) = 1/((k+1).(4k+3)).

If you are satisfied with the absence of 1/c you are ready:
4/(4k+3) = 1/(k+1) + 1/((k+1).(4k+3))

Otherwise, the problem is to write any of the two fractions at the right hand side as the sum of two other - different - Egyptian fractions. You can use

1/N = 1/(N+1) + 1/(N.(N+1)) for any N giving two schemes:

for N = k+1 we get
4/(4k+3) = 1/(k+2) + 1/((k+1).(k+2)) + 1/((k+1).(4k+3)) and

for N=(k+1).(4k+3) we get
4/(4k+3) = 1/(k+1) + 1/((k+1).(4k+3) + 1) + 1/((k+1).(4k+3).((k+1).(4k+3) + 1)).

Since by the same method, we can easily find solutions for even n, the real problem is in the case that n = 1 mod 4. For all but some cases, the Erdös-Straus conjecture has been proved.

Except for the case where n = 1^2, 11^2, 13^2, 17^2, 19^2, or 23^2 mod 840. Note that these are all subcases of the case n = 1 mod 4.

See p.34 of http://www.kevingong.com/Math/EgyptianFractions.pdf for conclusion and more references.

7. Aug 22, 2005

### VietDao29

Please recheck the problem... For n = 1:
$$4 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 3$$
$$\Leftrightarrow 4 \leq 3$$!!!!!
So it's wrong for n = 1.