# Difficulty with the Erdos-Straus Conjecture

• Descender
In summary, the Erdos-Straus Conjecture proposes that for n >= 1, 4/n = 1/a + 1/b + 1/c has positive integer solutions. The conjecture provides a pattern for n = 2k (evens), n = 4k+3 (odds), and n = 4k+1 (odds), but the pattern for n = 4k+3 has not yet been found. Several mathematicians have made progress in solving the conjecture, but it has yet to be proven.

#### 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.

Originally posted by 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.
n>=1
n=2k
it should be 2k>=1 not k>=1

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

but I am not sure this what you are after.

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.

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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.

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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.

Descender said:
The Erdos-Straus Conjecture proposes that:

For n >= 1, 4/n = 1/a + 1/b + 1/c, has postive integer solutions.
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.
For n > 1, 4/n = 1/a + 1/b + 1/c, has postive integer solutions.
Viet Dao,

## 1. What is the Erdos-Straus Conjecture?

The Erdos-Straus Conjecture is a mathematical conjecture proposed by mathematicians Paul Erdos and Ernst Straus in 1948. It states that every positive integer greater than 2 can be written as the sum of three distinct fractions, each with a denominator of the form (k+1)(k+2) for some positive integer k.

## 2. Is the Erdos-Straus Conjecture proven?

No, the Erdos-Straus Conjecture has not been proven. It is still an open problem in mathematics and remains unproven despite decades of research.

## 3. Why is the Erdos-Straus Conjecture important?

The Erdos-Straus Conjecture is important because it is a fundamental problem in number theory and has connections to other areas of mathematics. It has also sparked interest and research in related problems and has led to the discovery of new mathematical techniques.

## 4. What progress has been made towards solving the Erdos-Straus Conjecture?

While the Erdos-Straus Conjecture remains unproven, progress has been made towards understanding the problem. Some special cases have been proven and there have been various approaches and techniques proposed by mathematicians to try and solve the conjecture.

## 5. What is the significance of the Erdos-Straus Conjecture in modern mathematics?

The Erdos-Straus Conjecture is significant in modern mathematics as it is an unsolved problem that has captured the interest of mathematicians for many years. It continues to be a subject of study and research, and its eventual resolution could potentially lead to important developments in number theory and related fields.