Erdős–Straus conjecture

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The Erdős–Straus conjecture states that for all integers n ≥ 2, the rational number 4/n can be expressed as the sum of three unit fractions. That is, there exist positive integers x, y, and z such that

$\frac4n = \frac1x + \frac1y + \frac1z.$

The sum of these unit fractions is an Egyptian fraction representation of the number 4/n. For instance, for n = 1801, there exists a solution with x = 451, y = 295364, and z = 3249004:

$\frac4{1801} = \frac1{451} + \frac1{295364} + \frac1{3249004}.$

If we multiply both sides of this equation by nxyz we find an equivalent form 4xyz=n(xy+xz+yz) for the problem as a Diophantine equation. The restriction that x, y, and z be positive is essential to the difficulty of the problem, for if negative values were allowed the problem could be solved trivially via one of the two identities 4/(4k+1) = 1/k - 1/k(4k+1) and 4/(4k-1) = 1/k + 1/k(4k-1).

If n is a composite number, n = pq, then an expansion for 4/n could be found immediately from an expansion for 4/p or 4/q. Therefore, if a counterexample to the Erdős–Straus conjecture exists, the smallest n forming a counterexample would have to be a prime number.

Paul Erdős and Ernst G. Straus formulated the conjecture in 1948 (see, e.g., Elsholtz) but the earliest published reference to it appears to be a 1950 paper of Erdős.

1 Modular identities
2 Computational verification
3 Generalizations
4 See also
5 References
6 External links

[edit] Modular identities

The Hasse principle for Diophantine equations asserts that, to solve such an equation over the integers, we should combine solutions obtained modulo each possible prime number. On the face of it this principle makes little sense for the Erdős–Straus conjecture, as the equation 4xyz=n(xy+xz+yz) is easily solvable modulo any prime. Nevertheless, modular identities have proven a very important tool in the study of the conjecture.

For values of n satisfying certain congruence relations, we can find an expansion for 4/n automatically as an instance of a polynomial identity. For instance, whenever n ≡ 2 (mod 3), the equation is solvable, because 4/(3x+2) = 1/(3x+2) + 1/(x+1) + 1/(x+1)(3x+2) for all x, and plugging in x = (n-2)/3 gives the desired expansion for 4/n. A greedy algorithm finds a solution in three or fewer terms whenever n is not 1 or 17 (mod 24), and the n ≡ 17 (mod 24) case is covered by the 2 (mod 3) relation, so the only values of n for which these two methods do not find expansions in three or fewer terms are those congruent to 1 (mod 24).

If we could find sufficiently many such congruences to cover all values of n, the problem would be solved. However, as Mordell showed, an identity of this form for values of n congruent to r mod p can exist only when r is not a quadratic residue modulo p. Mordell lists identities of this form for any n that is 2 mod 3 (above), 3 mod 4, 5 mod 8, 2 or 3 mod 5, or 3, 5, or 6 mod 7, covering all the numbers that are not quadratic residues for those bases. However, for larger bases, not all nonresidues are known to be covered by relations of this type.

From Mordell's identities one can conclude that there exists a solution for all n except possibly those that are 1, 121, 169, 289, 361, or 529 modulo 840. 1009 is the smallest prime of this form. By combining larger classes of modular identities, Webb and others showed that the fraction of n in the interval [1,N] that can be counterexamples to the conjecture tends to zero in the limit as N goes to infinity.

Despite Mordell's result limiting the form these congruence identities can take, there is still some hope of using modular identities to prove the Erdős–Straus conjecture. No prime number can be a square, so by the Hasse-Minkowski theorem, whenever p is prime, there exists a larger prime q such that p is not a quadratic residue modulo q. One possible approach to proving the conjecture would be to find for each prime p a larger prime q and a congruence solving the 4/n problem for n ≡ p (mod q); if this could be done, no prime p could be a counterexample to the conjecture and the conjecture would be true.

[edit] Computational verification

Various authors have used computers to perform brute-force searches for counterexamples to the conjecture. These searches can be greatly sped up by considering only prime numbers that are not covered by known congruence relations. Searches of this type by Allan Swett confirmed that the conjecture is true for all n up to 10¹⁴.[1]

The number of distinct solutions to the 4/n problem, as a function of n, has also been found by computer searches for small n and appears to grow somewhat irregularly with n. Starting with n = 3, the numbers of distinct solutions with distinct denominators are

1, 1, 2, 5, 5, 6, 4, 9, 7, 15, 4, 14, 33, 22, 4, 21, 9, ... (sequence A073101 in OEIS).

Even for larger n there can be relatively few solutions; for instance there are only seven distinct solutions for n = 73.

[edit] Generalizations

A generalized version of the conjecture due to Wacław Sierpiński states that, for any positive k there exists a number N such that, for all n ≥ N, there exists a solution in positive integers to k/n = 1/x + 1/y + 1/z.

Hagedorn proved a related conjecture of R. H. Hardin and Neil Sloane that, for odd positive n, the equation 3/n = 1/x + 1/y + 1/z is always solvable with x, y, and z distinct, odd, and positive.

[edit] See also

Conjectures by Erdős

[edit] References

Elsholtz, Christian (2001), “Sums of k unit fractions”, Transactions of the American Mathematical Society 353 (8): 3209–3227, MR 1828604, DOI 10.1090/S0002-9947-01-02782-9 .

Erdős, Paul (1950), “Az 1/x₁ + 1/x₂ + ... + 1/x_n = a/b egyenlet egész számú megoldásairól (On a Diophantine Equation)”, Mat. Lapok. 1: 192–210, MR 0043117 .