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# Erdős–Straus conjecture

The Erdős–Straus conjecture concerns a Diophantine equation, refered to as the Erdős–Straus Diophantine equation, involving unit fractions.

Conjecture. (Erdős Pál, Ernst G. Straus) The equation[1]
${\displaystyle {\frac {4}{n}}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}},\quad 1\leq a\leq b\leq c,}$

always has solutions for any integer ${\displaystyle n\geq 2}$.

It is trivial to see that for ${\displaystyle n=1}$ there can't be any solution!

You only have to check for primes ${\displaystyle p}$, since any positive integer (except the unit 1) is a multiple ${\displaystyle m}$ of some prime ${\displaystyle p}$, and

${\displaystyle {\frac {4}{mp}}={\frac {1}{ma}}+{\frac {1}{mb}}+{\frac {1}{mc}},\quad 1\leq a\leq b\leq c,\quad m\geq 1.}$

Thus ${\displaystyle f(mp)\geq f(p)}$, where ${\displaystyle f(n)}$ is the number of solutions.

As of 2014, the conjecture has been checked up to 10 17.[2]

The Erdős–Straus conjecture means that we can always find some positive integer ${\displaystyle k}$ for which there exists at least one partition ${\displaystyle (a_{1},a_{2},a_{3})}$ of ${\displaystyle 4k}$ such that each of the three parts divides into ${\displaystyle nk}$, i.e.

${\displaystyle {\frac {4}{n}}={\frac {4k}{nk}}={\frac {a_{1}+a_{2}+a_{3}}{nk}}=\left({\tfrac {nk}{a_{1}}}\right)^{-1}+\left({\tfrac {nk}{a_{2}}}\right)^{-1}+\left({\tfrac {nk}{a_{3}}}\right)^{-1}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}},\quad a_{1}+a_{2}+a_{3}=4k,\quad a_{1}\,|\,nk,\,a_{2}\,|\,nk,\,a_{3}\,|\,nk,\quad k\geq 1.}$

Thus ${\displaystyle nk={\rm {lcm}}(a,b,c)}$, and

${\displaystyle a_{1}={\frac {{\rm {lcm}}(a,b,c)}{a}},\,a_{2}={\frac {{\rm {lcm}}(a,b,c)}{b}},\,a_{3}={\frac {{\rm {lcm}}(a,b,c)}{c}}.}$

Again, we only have to consider the primes. For example:

${\displaystyle {\frac {4}{5}}={\frac {4\cdot 2}{5\cdot 2}}={\frac {8}{5\cdot 2}}={\frac {5+2+1}{5\cdot 2}}={\frac {1}{2}}+{\frac {1}{5}}+{\frac {1}{10}}}$
${\displaystyle {\frac {4}{7}}={\frac {4\cdot 4}{7\cdot 4}}={\frac {16}{7\cdot 2^{2}}}={\frac {7+7+2}{7\cdot 2^{2}}}={\frac {1}{4}}+{\frac {1}{4}}+{\frac {1}{14}}}$
${\displaystyle {\frac {4}{11}}={\frac {4\cdot 4}{11\cdot 4}}={\frac {16}{11\cdot 2^{2}}}={\frac {11+4+1}{11\cdot 2^{2}}}={\frac {1}{4}}+{\frac {1}{11}}+{\frac {1}{44}}}$
${\displaystyle {\frac {4}{13}}={\frac {4\cdot 4}{13\cdot 4}}={\frac {16}{13\cdot 2^{2}}}={\frac {13+2+1}{13\cdot 2^{2}}}={\frac {1}{4}}+{\frac {1}{26}}+{\frac {1}{52}}}$

## Truth of the conjecture for primes

### Truth of the conjecture for 2

For the only even prime, i.e. 2, we have the solution

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

Thus, the conjecture is true for any positive even integer.

### Truth of the conjecture for primes congruent to 3 (mod 4)

If ${\displaystyle p\equiv 3{\pmod {4}}}$, we always have solutions since

${\displaystyle {\frac {4}{p}}={\frac {4\left({\tfrac {p+1}{2}}\right)}{p\left({\tfrac {p+1}{2}}\right)}}={\frac {1}{\left({\tfrac {p+1}{2}}\right)}}{\frac {2(p+1)}{p}}={\frac {1}{\left({\tfrac {p+1}{2}}\right)}}\left(2+{\frac {2}{p}}\right)={\frac {1}{\left({\tfrac {p+1}{2}}\right)}}\left(2+{\frac {1}{p}}+{\frac {1}{p}}\right)={\frac {1}{\left({\tfrac {p+1}{4}}\right)}}+{\frac {1}{\left({\tfrac {p+1}{2}}\right)p}}+{\frac {1}{\left({\tfrac {p+1}{2}}\right)p}}={\frac {1}{\left({\tfrac {p+1}{4}}\right)}}+{\frac {1}{t_{p}}}+{\frac {1}{t_{p}}},}$

where ${\displaystyle t_{p}}$ is the ${\displaystyle p}$th triangular number.

Thus, the conjecture is true for any positive integer divisible by a prime ${\displaystyle p}$ congruent to 3 (mod 4).

### Truth of the conjecture for primes congruent to 1 (mod 4)

If the conjecture is true for ${\displaystyle p\equiv 1{\pmod {4}}}$, then the conjecture would be true for any positive integer divisible by a prime congruent to 1 (mod 4).

## Erdős–Straus conjecture for unit fractions from the open unit interval

Equivalently, any unit fraction from the open unit interval, i.e. ${\displaystyle 0<{\tfrac {1}{n}}<1}$, is one quarter of the sum of three (distinct or not) positive unit fractions[3]

${\displaystyle {\frac {1}{n}}={\frac {1}{4}}\left({\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}\right),\quad 1\leq a\leq b\leq c.}$

## Erdős–Straus conjecture for prime unit fractions from the open unit interval

If a solution is known for ${\displaystyle n}$, then solutions are known for all multiples ${\displaystyle mn,\,m\geq 1,}$ since

${\displaystyle {\frac {1}{mn}}={\frac {1}{4}}\left({\frac {1}{ma}}+{\frac {1}{mb}}+{\frac {1}{mc}}\right),\quad 1\leq a\leq b\leq c.}$

For example, since

${\displaystyle {\frac {4}{7}}={\frac {2\cdot 3\cdot 4}{2\cdot 3\cdot 7}}={\frac {3\cdot 7+2+1}{2\cdot 3\cdot 7}}={\frac {2\cdot 7+7+3}{2\cdot 3\cdot 7}},}$

we have

${\displaystyle {\frac {4}{7}}={\frac {1}{2}}+{\frac {1}{21}}+{\frac {1}{42}}={\frac {1}{3}}+{\frac {1}{6}}+{\frac {1}{14}}.}$

We can then obtain solutions for ${\displaystyle n=42}$ simply by multiplying both sides by ${\displaystyle {\tfrac {1}{6}}}$:

${\displaystyle {\frac {4}{42}}={\frac {2}{21}}={\frac {1}{12}}+{\frac {1}{126}}+{\frac {1}{252}}={\frac {1}{18}}+{\frac {1}{36}}+{\frac {1}{84}}.}$

Thus it is sufficient to prove that the conjecture is true for prime unit fractions.

Any prime unit fraction from the open unit interval, i.e. ${\displaystyle 0<{\tfrac {1}{p}}<1}$, is one quarter of the sum of three (distinct or not) positive unit fractions[4]

${\displaystyle {\frac {1}{p}}={\frac {1}{4}}\left({\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}\right),\quad 1\leq a\leq b\leq c,}$

for any prime ${\displaystyle p}$.

## Solutions (a, b, c) of 4/n = 1/a + 1/b + 1/c

Distinct solutions (a, b, c) in lexicographic order of 4/n = 1/a + 1/b + 1/c for n > 1
${\displaystyle 4/n\,}$ #
4 / 2 (1, 2, 2) 1
4 / 3 (1, 4, 12), (1, 6, 6), (2, 2, 3) 3
4 / 4 (2, 3, 6), (2, 4, 4), (3, 3, 3) 3
4 / 5 (2, 4, 20), (2, 5, 10) 2
4 / 6 (2, 7, 42), (2, 8, 24), (2, 9, 18), (2, 10, 15), (2, 12, 12), (3, 4, 12), (3, 6, 6), (4, 4, 6) 8
4 / 7 (2, 15, 210), (2, 16, 112), (2, 18, 63), (2, 21, 42), (2, 28, 28), (3, 6, 14), (4, 4, 14) 7
4 / 8 (3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12), (4, 5, 20), (4, 6, 12), (4, 8, 8), (5, 5, 10), (6, 6, 6) 10

### Number of solutions (a, b, c) of 4/n = 1/a + 1/b + 1/c

A192786 Number of solutions of 4/n = 1/a + 1/b + 1/c in positive integers, n >= 1. (Unit fractions may be repeated.)

{0, 3, 12, 16, 12, 45, 36, 58, 36, 75, 48, 136, 24, 105, 240, 190, 24, 159, 66, 250, 186, 153, 132, 364, 78, 129, 180, 292, 42, 531, 114, 490, 198, 159, 426, 526, 60, 201, 450, ...}

A192787 Number of distinct solutions of 4/n = 1/a + 1/b + 1/c in positive integers satisfying 1 <= a <= b <= c, n >= 1. (Unit fractions may be repeated.)

{0, 1, 3, 3, 2, 8, 7, 10, 6, 12, 9, 21, 4, 17, 39, 28, 4, 26, 11, 36, 29, 25, 21, 57, 10, 20, 29, 42, 7, 81, 19, 70, 31, 25, 65, 79, 9, 32, 73, 96, 7, 86, 14, 62, 93, 42, 34, ...}

A073101 Number of solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z, n >= 2. (Unit fractions must be distinct.)

{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, 77, 12, 56, 87, 39, 32, 142, ...}

A?????? Number of distinct solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z, n >= 2. (Unit fractions must be distinct.)

{0, 1, ...}

## Solutions (a, b, c) of 4/p = 1/a + 1/b + 1/c

Distinct solutions (a, b, c) in lexicographic order of 4/p = 1/a + 1/b + 1/c for p prime
${\displaystyle 4/p\,}$ #
4 / 2 (1, 2, 2) 1
4 / 3 (1, 4, 12), (1, 6, 6), (2, 2, 3) 3
4 / 5 (2, 4, 20), (2, 5, 10) 2
4 / 7 (2, 15, 210), (2, 16, 112), (2, 18, 63), (2, 21, 42), (2, 28, 28), (3, 6, 14), (4, 4, 14) 7
4 / 11 (3, 34, 1122), (3, 36, 396), (3, 42, 154), (3, 44, 132), (3, 66, 66), (4, 9, 396), (4, 11, 44), (4, 12, 33), (6, 6, 33) 9
4 / 13 (4, 18, 468), (4, 20, 130), (4, 26, 52), (5, 10, 130) 4
4 / 17 (5, 30, 510), (5, 34, 170), (6, 15, 510), (6, 17, 102) 4
4 / 19 (5, 96, 9120), (5, 100, 1900), (5, 114, 570), (5, 120, 456), (5, 190, 190), (6, 23, 2622), (6, 24, 456), (6, 30, 95), (6, 38, 57), (8, 12, 456), (10, 10, 95) 11
4 / 23 (6, 139, 19182), (6, 140, 9600), (6, 141, 6486), (6, 142, 4899), (6, 144, 3312), (6, 147, 2254), (6, 150, 1725), (6, 156, 1196), (6, 161, 966), (6, 174, 667), (6, 184, 552), (6, 207, 414), (6, 230, 345), (6, 276, 276), (7, 42, 138), (8, 23, 184), (8, 24, 138), (9, 16, 3312), (9, 18, 138), (10, 15, 138), (12, 12, 138) 21
4 / 29 (8, 78, 9048), (8, 80, 2320), (8, 87, 696), (8, 88, 638), (8, 116, 232), (10, 29, 290), (11, 22, 638) 7

### Number of solutions (a, b, c) of 4/p = 1/a + 1/b + 1/c

In 2012, Christian Elsholtz & Terence Tao established that

${\displaystyle N\log ^{2}N\ll \sum _{p\leq N}f(p)\ll N\log ^{2}N\log \log N,}$

where ${\displaystyle p}$ is a prime and ${\displaystyle f(p)}$ is the number of solutions to the Erdős–Straus Diophantine equation.

A192788 Number of solutions of 4/p = 1/a + 1/b + 1/c in positive integers, where p is the n-th prime. (Unit fractions may be repeated.)

{3, 12, 12, 36, 48, 24, 24, 66, 132, 42, 114, 60, 48, 84, 216, 90, 168, 72, 108, 246, 42, 228, 162, 66, 48, 102, 156, 150, 96, 84, 198, 192, 108, 222, 114, 192, 144, 144, ...}

A192789 Number of distinct solutions of 4/p = 1/a + 1/b + 1/c in positive integers, where p is the n-th prime. (Unit fractions may be repeated.)

{1, 3, 2, 7, 9, 4, 4, 12, 23, 7, 20, 10, 8, 15, 37, 15, 29, 12, 19, 42, 7, 39, 28, 11, 8, 17, 27, 26, 16, 14, 34, 33, 18, 38, 19, 33, 24, 25, 68, 27, 52, 18, 69, 6, 25, 43, 32, ...}

A?????? Number of solutions (x,y,z) to 4/p = 1/x + 1/y + 1/z satisfying 0 < x < y < z, where p is the n-th prime. (Unit fractions must be distinct.)

{0, 1, 2, 5, 7, 4, 4, 9, 19, 7, 17, 9, 12, 32, ...}

A?????? Number of distinct solutions (x,y,z) to 4/p = 1/x + 1/y + 1/z satisfying 0 < x < y < z, where p is the n-th prime. (Unit fractions must be distinct.)

{0, 1, ...}

## Notes

1. A more constrained variant of the conjecture, where the unit fractions must be distinct, is
${\displaystyle {\frac {4}{n}}={\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}},\quad 0
always has solutions for any integer ${\displaystyle n\geq 3}$. (Note that there is no solution for ${\displaystyle n=2}$ when we require distinct unit fractions.)
2. https://arxiv.org/abs/1406.6307
3. Equivalent conjecture: any nonunit positive integer, i.e. greater than 1, is ${\displaystyle {\tfrac {4}{3}}}$ times the harmonic mean of three (distinct or not) positive unit fractions
${\displaystyle n={\frac {4}{3}}\left({\frac {{\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}{3}}\right)^{-1},\quad 1\leq a\leq b\leq c,}$
for any positive integer ${\displaystyle n>1}$.
4. Equivalent conjecture: any prime is ${\displaystyle {\tfrac {4}{3}}}$ times the harmonic mean of three (distinct or not) positive unit fractions
${\displaystyle p={\frac {4}{3}}\left({\frac {{\frac {1}{a}}+{\frac {1}{b}}+{\frac {1}{c}}}{3}}\right)^{-1},\quad 1\leq a\leq b\leq c,}$
for any prime ${\displaystyle p}$. (If it's true for any prime ${\displaystyle p}$, then it's true for any ${\displaystyle kp,\,k\geq 1,}$ thus for any nonunit positive integer, i.e. greater than 1, the unit 1 being the empty product of primes.)