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Erdős–Straus conjecture

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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]

always has solutions for any integer .

It is trivial to see that for there can't be any solution!

You only have to check for primes , since any positive integer (except the unit 1) is a multiple of some prime , and

Thus , where is the number of solutions.

As of 2011, the conjecture have been checked up to 10 14, for 2 and all primes congruent to 1 (mod 4). (We don't need to check for primes congruent to 3 (mod 4), see the proof below.)

The Erdős–Straus conjecture means that we can always find some positive integer for which there exists at least one partition of such that each of the three parts divides into , i.e.

Thus , and

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

Truth of the conjecture for primes

Truth of the conjecture for 2

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

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

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

If , we always have solutions since

where is the th triangular number.

Thus, the conjecture is true for any positive integer divisible by a prime congruent to 3 (mod 4).

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

If the conjecture is true for , 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. , is one quarter of the sum of three (distinct or not) positive unit fractions[2]

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

If a solution is known for , then solutions are known for all multiples since

For example, since

we have

We can then obtain solutions for simply by multiplying both sides by :

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. , is one quarter of the sum of three (distinct or not) positive unit fractions[3]

for any prime .

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
#
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
#
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

where is a prime and 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
    always has solutions for any integer . (Note that there is no solution for when we require distinct unit fractions.)
  2. Equivalent conjecture: any nonunit positive integer, i.e. greater than 1, is times the harmonic mean of three (distinct or not) positive unit fractions
    for any positive integer .
  3. Equivalent conjecture: any prime is times the harmonic mean of three (distinct or not) positive unit fractions
    for any prime . (If it's true for any prime , then it's true for any thus for any nonunit positive integer, i.e. greater than 1, the unit 1 being the empty product of primes.)

External links

  • Christian Elsholtz & Terence Tao, "Counting the number of solutions to the Erdos-Straus equation on unit fractions," arXiv:1107.1010.