Euler's sum of powers conjecture

Euler's conjecture is a disproved conjecture in mathematics related to Fermat's Last Theorem. It was proposed by Leonhard Euler in 1769. It states that for all integers n and k greater than 1, if the sum of n many kth powers of positive integers is itself a kth power, then n is greater than or equal to k:

a k
1  + a k
2  + ... + a k
n  = b^k ⇒ n ≥ k

The conjecture represents an attempt to generalize Fermat's Last Theorem, which is the special case n = 2: if a k
1  + a k
2  = b^k, then 2 ≥ k.

Although the conjecture holds for the case k = 3 (which follows from Fermat's Last Theorem for the third powers), it was disproved for k = 4 and k = 5. It is unknown whether the conjecture fails or holds for any value k ≥ 6.

Background[]

Euler was aware of the equality 59⁴ + 158⁴ = 133⁴ + 134⁴ involving sums of four fourth powers; this however is not a counterexample because no term is isolated on one side of the equation. He also provided a complete solution to the four cubes problem as in Plato's number 3³ + 4³ + 5³ = 6³ or the taxicab number 1729.^[1][2] The general solution of the equation

${\displaystyle x_{1}^{3}+x_{2}^{3}=x_{3}^{3}+x_{4}^{3}}$

is

${\displaystyle x_{1}=1-(a-3b)(a^{2}+3b^{2}),\quad x_{2}=(a+3b)(a^{2}+3b^{2})-1}$

${\displaystyle x_{3}=(a+3b)-(a^{2}+3b^{2})^{2},\quad x_{4}=(a^{2}+3b^{2})^{2}-(a-3b)}$

where a and b are any integers.

Counterexamples[]

Euler's conjecture was disproven by and in 1966 when, through a direct computer search on a CDC 6600, they found a counterexample for k = 5.^[3] This was published in a paper comprising just two sentences.^[3] A total of three primitive (that is, in which the summands do not all have a common factor) counterexamples are known:

27⁵ + 84⁵ + 110⁵ + 133⁵ = 144⁵ (Lander & Parkin, 1966),

(−220)⁵ + 5027⁵ + 6237⁵ + 14068⁵ = 14132⁵ (Scher & Seidl, 1996), and

55⁵ + 3183⁵ + 28969⁵ + 85282⁵ = 85359⁵ (Frye, 2004).

In 1988, Noam Elkies published a method to construct an infinite sequence of counterexamples for the k = 4 case.^[4] His smallest counterexample was

2682440⁴ + 15365639⁴ + 18796760⁴ = 20615673⁴.

A particular case of Elkies' solutions can be reduced to the identity^[5][6]

(85v² + 484v − 313)⁴ + (68v² − 586v + 10)⁴ + (2u)⁴ = (357v² − 204v + 363)⁴

where

u² = 22030 + 28849v − 56158v² + 36941v³ − 31790v⁴.

This is an elliptic curve with a rational point at v₁ = −31/467. From this initial rational point, one can compute an infinite collection of others. Substituting v₁ into the identity and removing common factors gives the numerical example cited above.

In 1988, found the smallest possible counterexample

95800⁴ + 217519⁴ + 414560⁴ = 422481⁴

for k = 4 by a direct computer search using techniques suggested by Elkies. This solution is the only one with values of the variables below 1,000,000.^[7]

Generalizations[]

In 1967, L. J. Lander, T. R. Parkin, and John Selfridge conjectured^[8] that if

${\displaystyle \sum _{i=1}^{n}a_{i}^{k}=\sum _{j=1}^{m}b_{j}^{k}}$,

where a_i ≠ b_j are positive integers for all 1 ≤ i ≤ n and 1 ≤ j ≤ m, then m + n ≥ k. In the special case m = 1, the conjecture states that if

${\displaystyle \sum _{i=1}^{n}a_{i}^{k}=b^{k}}$

(under the conditions given above) then n ≥ k − 1.

The special case may be described as the problem of giving a partition of a perfect power into few like powers. For k = 4, 5, 7, 8 and n = k or k − 1, there are many known solutions. Some of these are listed below. As of 2002, there are no solutions for ${\displaystyle k=6}$ whose final term is ≤ 730000.^[9]

See OEIS: A347773 for more data.

k = 3[]

3³ + 4³ + 5³ = 6³ (Plato's number 216)

This is the case a = 1, b = 0 of Srinivasa Ramanujan's formula

${\displaystyle (3a^{2}+5ab-5b^{2})^{3}+(4a^{2}-4ab+6b^{2})^{3}+(5a^{2}-5ab-3b^{2})^{3}=(6a^{2}-4ab+4b^{2})^{3}.}$ ^[10]

A cube as the sum of three cubes can also be parameterized as

${\displaystyle a^{3}(a^{3}+b^{3})^{3}=b^{3}(a^{3}+b^{3})^{3}+a^{3}(a^{3}-2b^{3})^{3}+b^{3}(2a^{3}-b^{3})^{3}}$

or as

${\displaystyle a^{3}(a^{3}+2b^{3})^{3}=a^{3}(a^{3}-b^{3})^{3}+b^{3}(a^{3}-b^{3})^{3}+b^{3}(2a^{3}+b^{3})^{3}.}$^[10]

The number 2 100 000³ can be expressed as the sum of three cubes in nine different ways.^[10]

k = 4[]

95800⁴ + 217519⁴ + 414560⁴ = 422481⁴ (R. Frye, 1988)^[4]

30⁴ + 120⁴ + 272⁴ + 315⁴ = 353⁴ (R. Norrie, 1911)^[8]

This is the smallest solution to the problem by R. Norrie.

k = 5[]

27⁵ + 84⁵ + 110⁵ + 133⁵ = 144⁵ (Lander & Parkin, 1966)^[11][12][13]

19⁵ + 43⁵ + 46⁵ + 47⁵ + 67⁵ = 72⁵ (Lander, Parkin, Selfridge, smallest, 1967)^[8]

21⁵ + 23⁵ + 37⁵ + 79⁵ + 84⁵ = 94⁵ (Lander, Parkin, Selfridge, second smallest, 1967)^[8]

7⁵ + 43⁵ + 57⁵ + 80⁵ + 100⁵ = 107⁵ (Sastry, 1934, third smallest)^[8]

k = 7[]

127⁷ + 258⁷ + 266⁷ + 413⁷ + 430⁷ + 439⁷ + 525⁷ = 568⁷ (M. Dodrill, 1999)^[14]

k = 8[]

90⁸ + 223⁸ + 478⁸ + 524⁸ + 748⁸ + 1088⁸ + 1190⁸ + 1324⁸ = 1409⁸ (S. Chase, 2000)^[15]

See also[]

• Jacobi–Madden equation
• Prouhet–Tarry–Escott problem
• Beal's conjecture
• Pythagorean quadruple
• Generalized taxicab number
• Sums of powers, a list of related conjectures and theorems

References[]

External links[]

• Tito Piezas III, A Collection of Algebraic Identities
• Jaroslaw Wroblewski, Equal Sums of Like Powers
• Ed Pegg Jr., Math Games, Power Sums
• James Waldby, A Table of Fifth Powers equal to a Fifth Power (2009)
• , J.-C. Meyrignac, U. Beckert, All solutions of the Diophantine equation a⁶ + b⁶ = c⁶ + d⁶ + e⁶ + f⁶ + g⁶ for a,b,c,d,e,f,g < 250000 found with a distributed Boinc project
• EulerNet: Computing Minimal Equal Sums Of Like Powers
• Weisstein, Eric W. "Euler's Sum of Powers Conjecture". MathWorld.
• Weisstein, Eric W. "Euler Quartic Conjecture". MathWorld.
• Weisstein, Eric W. "Diophantine Equation--4th Powers". MathWorld.
• Euler's Conjecture at library.thinkquest.org
• A simple explanation of Euler's Conjecture at Maths Is Good For You!