Erdős-Straus conjecture

The number-theoretical Erdős-Straus conjecture (after the mathematicians Paul Erdős and Ernst Gabor Straus ) says that there is always a sum of three positive fractions . It was established in 1948 and is one of Paul Erdős' many conjectures . ${\displaystyle {\frac {4}{n}}}$

The presumption

The equation has a solution for every natural , where , and are also natural numbers. ${\displaystyle {\frac {4}{n}}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}$ ${\displaystyle n\geq 2}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle c}$

It is immediately clear that four parent fractions are always sufficient (choose four times for the summands ). The conjecture corresponds to the next smaller number of stem fraction summands. ${\displaystyle {\frac {1}{n}}}$

There are two ways in which such a representation can be understood as the sum of three original fractions. One can be of the opinion that it does not matter whether certain denominators of the individual fractions are the same, but one can also be of the opinion that it does not matter whether certain denominators of the individual fractions are the same. For the Erdős-Straus conjecture, however, it makes no difference whether the same denominators are allowed or not: ${\displaystyle n\geq 3}$

If there is any representation of with three fractions, regardless of whether certain denominators are the same or not, then there is definitely also a representation with different denominators. You can always convert two original fractions with the same denominators into two original fractions with different denominators: ${\displaystyle {\frac {4}{n}}}$

${\displaystyle {\frac {1}{2r}}+{\frac {1}{2r}}={\frac {1}{r+1}}+{\frac {1}{r(r+1)}}\qquad }$ when the denominator is an even number

${\displaystyle {\frac {1}{2r+1}}+{\frac {1}{2r+1}}={\frac {1}{r+1}}+{\frac {1}{(r+1)(2r+1)}}\qquad }$ when the denominator is an odd number

So you can transform two original fractions with the same denominator into two original fractions with two different denominators. If you get two identical denominators again (together with the denominator of the third common fraction), you repeat this transformation until you have three common fractions with three different denominators.

Geometric interpretation

The geometric interpretation of the Erdős-Straus conjecture provides a cuboid with the edge lengths , and ( , and natural numbers) for every natural one , so that its 8-fold volume divided by its surface yields the value of units of length. ${\displaystyle n>1}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle c}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle c}$ ${\displaystyle n}$

Examples

• If there is a representation of with through three parent fractions, there is definitely at least one representation of as the sum of 3 parent fractions with different denominators. In this case , however, there is only one solution, namely ${\displaystyle {\frac {4}{n}}}$ ${\displaystyle n\geq 3}$ ${\displaystyle {\frac {4}{n}}}$ ${\displaystyle n=2}$

${\displaystyle {\frac {4}{2}}={\frac {1}{1}}+{\frac {1}{2}}+{\frac {1}{2}}}$

In this case, there is no way to get different denominators. In all other cases, the stem fractions are usually ordered according to the size of their denominator, so that applies. ${\displaystyle 0<a<b<c}$

• For the two cases or there is only one solution, namely ${\displaystyle n=3}$ ${\displaystyle n=4}$

${\displaystyle {\frac {4}{3}}={\frac {1}{1}}+{\frac {1}{4}}+{\frac {1}{12}}\qquad }$ and ${\displaystyle \qquad {\frac {4}{4}}={\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{6}}}$

• There are sometimes several solutions to this problem, such as for which there are two solutions: ${\displaystyle n=5}$

${\displaystyle {\frac {4}{5}}={\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{20}}={\frac {1}{2}}+{\frac {1}{5}}+{\frac {1}{10}}}$

• The next sequence of numbers indicates how many display options there are for in the form with (with ascending ) ${\displaystyle {\frac {4}{n}}}$ ${\displaystyle {\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}$ ${\displaystyle 0<a<b<c}$ ${\displaystyle n=1,2,3,\ldots }$

0, 0, 1, 1, 2, 5, 5, 6, 4, 9, 7, 15, 4, 14, 33, 22, 4, 21, 9, 30, 25, 22, 19, 45, 10, 17, 25, 36, 7, 72, 17, 62, 27, 22, 59, 69, 9, 29, 67, 84, 7, ... (sequence A073101 in OEIS )

Example 1:

At the n = 8th position of this list there is 6. This means that there are 6 possibilities to represent the fraction with three original fractions. In particular, they are the following: ${\displaystyle {\frac {4}{8}}}$

${\displaystyle {\frac {4}{8}}\quad ={\frac {1}{3}}+{\frac {1}{7}}+{\frac {1}{42}}\quad ={\frac {1}{3}}+{\frac {1}{8}}+{\frac {1}{24}}\quad ={\frac {1}{3}}+{\frac {1}{9}}+{\frac {1}{18}}\quad ={\frac {1}{3}}+{\frac {1}{10}}+{\frac {1}{15}}\quad ={\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{20}}\quad ={\frac {1}{4}}+{\frac {1}{6}}+{\frac {1}{12}}}$

(Note: a good online fraction calculator for small ones can be found on) ${\displaystyle n}$

Example 2:

The number of different ways to represent a fraction as the sum of three original fractions can be very high (in relation to ), but also quite low. For example, the fraction ( i.e. for ) can be represented in exactly 4914 different ways as the sum of three original fractions. In contrast, the fraction (i.e. for ) can only be represented in 27 different ways as the sum of three original fractions. ${\displaystyle {\frac {4}{n}}}$ ${\displaystyle n}$ ${\displaystyle {\frac {4}{960}}}$ ${\displaystyle n=960}$ ${\displaystyle {\frac {4}{937}}}$ ${\displaystyle n=937}$

Remarks

• If you multiply the equation by (remove the denominators), you get the following Diophantine equation : ${\displaystyle {\frac {4}{n}}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}$ ${\displaystyle nabc}$

${\displaystyle 4abc=n\cdot (bc+ac+ab)}$

Due to the fact that one is dealing with a Diophantine equation, one can apply the local-global principle popular in number theory : from the solvability of this Diophantine equation modulo all prime numbers , the solvability of the original equation can be inferred .

• For everyone with natural the claim is easy to prove with : ${\displaystyle n=4k}$ ${\displaystyle k\in \mathbb {N} }$ ${\displaystyle a=b=c=3k}$

${\displaystyle {\frac {4}{n}}={\frac {4}{4k}}={\frac {1}{k}}={\frac {3}{3k}}={\frac {1}{3k}}+{\frac {1}{3k}}+{\frac {1}{3k}}}$

• The more general case with the natural is also very easy to solve with and : ${\displaystyle n=2k}$ ${\displaystyle k\in \mathbb {N} }$ ${\displaystyle a=k}$ ${\displaystyle b=c=2k}$

${\displaystyle {\frac {4}{n}}={\frac {4}{2k}}={\frac {2}{2k}}+{\frac {1}{2k}}+{\frac {1}{2k}}={\frac {1}{k}}+{\frac {1}{2k}}+{\frac {1}{2k}}}$.

• In order to solve the Erdős-Straus conjecture, it is sufficient if, instead of only fractions of the form with prime investigated. The Erdős-Straus conjecture can thus be broken down into the following conjecture: ${\displaystyle {\frac {4}{n}}}$ ${\displaystyle {\frac {4}{p}}}$ ${\displaystyle p\in \mathbb {P} }$

The equation has for each prime number , a solution, wherein , and are natural numbers. ${\displaystyle {\frac {4}{p}}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}$ ${\displaystyle p\in \mathbb {P} }$ ${\displaystyle p>2}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle c}$

• If the equation with different pairs does not have an integral solution, then for every prime divisor of : ${\displaystyle {\frac {4}{n}}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}$ ${\displaystyle a,b,c}$ ${\displaystyle p}$ ${\displaystyle n}$

${\displaystyle p\equiv 1{\pmod {24}}}$

and therefore also applies:

${\displaystyle n\equiv 1{\pmod {24}}}$

It also follows that there is always a solution to the equation for everyone . In this case, the Erdős-Straus conjecture is solved. ${\displaystyle n\not \equiv 1{\pmod {24}}}$

• If there is actually a counterexample to the Erdős-Straus conjecture (i.e. if there is a fraction that can not be represented by three different ancestral fractions), it must meet one of the following six congruences : ${\displaystyle {\frac {4}{n}}}$ ${\displaystyle n}$

${\displaystyle n\equiv 1{\pmod {840}}}$

${\displaystyle n\equiv 11^{2}=121{\pmod {840}}}$

${\displaystyle n\equiv 13^{2}=169{\pmod {840}}}$

${\displaystyle n\equiv 17^{2}=289{\pmod {840}}}$

${\displaystyle n\equiv 19^{2}=361{\pmod {840}}}$

${\displaystyle n\equiv 23^{2}=529{\pmod {840}}}$

The first two prime numbers (and only these have to be tested, as already mentioned above), which fulfill one of the above six congruences, are and . You can see that you get quite high values ​​for relatively quickly and you don't have to test all of them by far . ${\displaystyle n=p_{1}=1009\equiv 169{\pmod {840}}}$ ${\displaystyle n=p_{2}=1201\equiv 361{\pmod {840}}}$ ${\displaystyle n}$ ${\displaystyle n}$

Further, more precise congruence conditions for are specified in FIG. ${\displaystyle n}$

• Allan Swett showed in 1999 that the Erdős-Straus conjecture is up to . Serge E. Salez extended this limit in 2014 . ${\displaystyle n=10^{14}}$ ${\displaystyle n=10^{17}}$

Decomposition into three not necessarily positive fractions

In the introduction it is mentioned that it is necessary that the three main fractions to be broken down into must be positive. If negative original fractions are also allowed, you can break down any fraction of the form into three original fractions and the conjecture would no longer be a conjecture, but a proven mathematical theorem . If negative fractions are allowed, there are at least two trivial ways to represent odd (for even , this conjecture was already proven in the previous section): ${\displaystyle {\frac {4}{n}}}$ ${\displaystyle {\frac {4}{n}}}$ ${\displaystyle n}$ ${\displaystyle n}$

Case 1: ${\displaystyle n=4k+1}$

${\displaystyle {\frac {4}{4k+1}}={\frac {1}{k}}-{\frac {1}{k(4k+1)}}={\frac {1}{2k}}+{\frac {1}{2k}}-{\frac {1}{k(4k+1)}}}$

Case 2: ${\displaystyle n=4k-1}$

${\displaystyle {\frac {4}{4k-1}}={\frac {1}{k}}+{\frac {1}{k(4k-1)}}={\frac {1}{2k}}+{\frac {1}{2k}}+{\frac {1}{k(4k-1)}}.}$

As a variant for odd there is also the following representation in three original fractions, but one of which is always negative: ${\displaystyle n}$

${\displaystyle {\frac {4}{n}}={\frac {1}{(n-1)/2}}+{\frac {1}{(n+1)/2}}-{\frac {1}{n(n-1)(n+1)/4}}.}$

Mini Erdős Straus conjecture

• A variant of the Erdős-Straus conjecture is the mini Erdős-Straus conjecture :

The equation has a solution for every natural , where and are natural numbers. ${\displaystyle {\frac {3}{n}}={\frac {1}{a}}+{\frac {1}{b}}}$ ${\displaystyle n\geq 2}$ ${\displaystyle a}$ ${\displaystyle b}$

This conjecture is wrong because there is no solution for natural and for the equation if and only if all prime factors of have the form . ${\displaystyle {\frac {3}{n}}={\frac {1}{a}}+{\frac {1}{b}}}$ ${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle n}$ ${\displaystyle 6k+1}$

Generalizations

Two conjectures follow, which supplement or generalize the Erdős-Straus conjecture:

• The Polish mathematician Wacław Sierpiński made the following amendment to the Erdős-Straus conjecture in 1956:

There is a number so the equation can be solved for all natural ones. ${\displaystyle N}$ ${\displaystyle {\frac {5}{n}}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}$ ${\displaystyle n\geq N}$

In other words: From a certain point on , all fractions of the form can be represented by three parent fractions. ${\displaystyle n}$ ${\displaystyle {\frac {5}{n}}}$

• The Polish mathematician Andrzej Schinzel generalized the Erdős-Straus conjecture as follows:

For any given natural number there is one , so the equation can be solved for all natural ones. ${\displaystyle m\in \mathbb {N} }$ ${\displaystyle N}$ ${\displaystyle {\frac {m}{n}}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}$ ${\displaystyle n\geq N}$

In other words: from a certain point on, all fractions of the form can be represented by three parent fractions. ${\displaystyle n}$ ${\displaystyle m}$ ${\displaystyle {\frac {m}{n}}}$

Web links

• Koichi Yamamoto: On the diophantine Equation 4 / n = 1 / x + 1 / y + 1 / z . Memoirs of the Faculty of Science, Kyushu University, Ser. A 19 (1), December 20, 1964, pp. 37-47 , accessed January 6, 2020 . 

• Gerd Hofmeister, Peter Stoll: Note on Egyptian fractions. Journal for pure and applied mathematics, January 9, 1985, pp. 141-145 , accessed January 7, 2020 . 
• Jamel Ghannouchi: The Erdos conjecture The proof. International Journal of Science and Research, January 19, 2015, p. 3 , accessed January 6, 2020 . 
• Eric W. Weisstein: Erdős-Straus Conjecture. Wolfram MathWorld, accessed January 6, 2020 . 

Individual evidence

1. ↑ Paul Erdős : Az 1 / x ₁ + 1 / x ₂ + ... + 1 / x _n = a / b egyenlet egész számú megoldásairól (On a Diophantine Equation). P. 210. Mat. Lapok. 1 , 1950, pp. 192–210 , accessed January 5, 2020 (Hungarian). 
2. ^ Christian Elsholtz: Sums of k unit fractions. Conjecture 1, here it is claimed that this presumption was first made in 1948. Transactions of the American Mathematical Society 353 (8), April 12, 2001, pp. 3209-3227 , accessed January 5, 2020 . 
3. ^ Algorithms for Egyptian Fractions - Theorem
4. ↑ Tanzô Takenouchi: On an Indeterminate Equation. Section (III), Solutions in which two x ’s are equal on page 80. Proc. Physico-Mathematical Soc. of Japan (3rd ser.) 3, April 3, 1921, pp. 78-92 , accessed January 5, 2020 . 
5. ↑ Arndt Brünner's fractions calculator
6. ↑ JW Sander: On 4 / n = 1 / x + 1 / y + 1 / z and Iwaniec 'Half Dimensional Sieve. Journal of Number Theory 46 (2), February 1994, pp. 123-136 , accessed January 5, 2020 . 
7. ^ Erdős-Straus conjecture. Mathematics, accessed January 5, 2020 . 
8. ↑ Konstantine Zelator: An ancient Egyptian problem: The diophantine equation 4 / n = 1 / x + 1 / y + 1 / z , n ≥ 2. Theorem December 2, 2009, pp. 1–9 , accessed on January 5, 2020 . 
9. ↑ ^{a b} Eugen J. Ionascu, Andrew Wilson: On the Erdös-Straus Conjecture. Theorem 2-4 on p. 24ff. Columbus State University, 2010, pp. 21-30 , accessed January 7, 2020 . 
10. ↑ Maria Monks, Ameya Velingker: On the Erdös-Straus Conjecture: Properties of Solutions to its underlying Diophantine Equation. P. 1 below. Pp. 1–21 , accessed January 8, 2020 . 
11. ↑ Louis J. Mordell : Diophantine Equations , Academic Press , 1967, pp. 287-290
12. ^ Serge E. Salez: The Erdős-Straus conjecture New modular equations and checking up to N = 10 ¹⁷ . June 24, 2014, pp. 1–13 , accessed January 5, 2020 . 
13. ↑ John H. Jaroma: On expanding 4 / n into three Egyptian fractions. Crux Mathematicorum 30 (1), 2004, pp. 36-37 , accessed January 5, 2020 . 
14. ^ Hans Humenberger : Egyptian fractions - Representations as sums of unit fractions. Section 3.3 on page 275f. Mathematics and Computer Education, pp. 268–283 , accessed January 5, 2020 . 
15. ↑ Algorithms for Egyptian Fractions, proof of theorem according to Numerator 3
16. ↑ Wacław Sierpiński : Sur les décompositions de nombres rationnels en fractions primaires , Mathesis 65 , 1956, pp. 16–32
17. ↑ ^{a b} Ji Hoon Chun: Egyptian fractions, Sylvester's sequence, and the Erdős-Straus conjecture. Section 1.6 on p. 6. August 1, 2011, pp. 1–7 , accessed on January 8, 2020 . 
18. ^ RC Vaughan: On a problem of Erdős, Straus and Schinzel , Mathematika 17 (02), 1970, pp. 193–198