Euler's sum of powers conjecture

In number theory, Euler's conjecture is a disproved conjecture 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:

$${\displaystyle a_{1}^{k}+a_{2}^{k}+\dots +a_{n}^{k}=b^{k}\implies n\geq k}$$

The conjecture represents an attempt to generalize Fermat's Last Theorem, which is the special case n = 2: if ${\displaystyle a_{1}^{k}+a_{2}^{k}=b^{k},}$ 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.

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

$${\displaystyle {\begin{aligned}x_{1}&=\lambda (1-(a-3b)(a^{2}+3b^{2}))\\[2pt]x_{2}&=\lambda ((a+3b)(a^{2}+3b^{2})-1)\\[2pt]x_{3}&=\lambda ((a+3b)-(a^{2}+3b^{2})^{2})\\[2pt]x_{4}&=\lambda ((a^{2}+3b^{2})^{2}-(a-3b))\end{aligned}}}$$

where a, b and ${\displaystyle {\lambda }}$ are any rational numbers.

Euler's conjecture was disproven by L. J. Lander and T. R. Parkin 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: $${\displaystyle {\begin{aligned}144^{5}&=27^{5}+84^{5}+110^{5}+133^{5}\\14132^{5}&=(-220)^{5}+5027^{5}+6237^{5}+14068^{5}\\85359^{5}&=55^{5}+3183^{5}+28969^{5}+85282^{5}\end{aligned}}}$$ (Lander & Parkin, 1966); (Scher & Seidl, 1996); (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 $${\displaystyle 20615673^{4}=2682440^{4}+15365639^{4}+18796760^{4}.}$$

A particular case of Elkies' solutions can be reduced to the identity^[5][6] $${\displaystyle (85v^{2}+484v-313)^{4}+(68v^{2}-586v+10)^{4}+(2u)^{4}=(357v^{2}-204v+363)^{4},}$$ where $${\displaystyle u^{2}=22030+28849v-56158v^{2}+36941v^{3}-31790v^{4}.}$$ 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, Roger Frye found the smallest possible counterexample $${\displaystyle 95800^{4}+217519^{4}+414560^{4}=422481^{4}}$$ 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]

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.

See OEIS: A347773 for more data.

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

This is the case a = 1, b = 0 of Srinivasa Ramanujan's formula^[9]

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

A cube as the sum of three cubes can also be parameterized in one of two ways:^[9]

$${\displaystyle {\begin{aligned}a^{3}(a^{3}+b^{3})^{3}&=b^{3}(a^{3}+b^{3})^{3}+a^{3}(a^{3}-2b^{3})^{3}+b^{3}(2a^{3}-b^{3})^{3}\\[6pt]a^{3}(a^{3}+2b^{3})^{3}&=a^{3}(a^{3}-b^{3})^{3}+b^{3}(a^{3}-b^{3})^{3}+b^{3}(2a^{3}+b^{3})^{3}\end{aligned}}}$$

The number 2 100 000³ can be expressed as the sum of three cubes in nine different ways.^[9]

$${\displaystyle {\begin{aligned}422481^{4}&=95800^{4}+217519^{4}+414560^{4}\\[4pt]353^{4}&=30^{4}+120^{4}+272^{4}+315^{4}\end{aligned}}}$$ (R. Frye, 1988);^[4] (R. Norrie, smallest, 1911).^[8]

$${\displaystyle {\begin{aligned}144^{5}&=27^{5}+84^{5}+110^{5}+133^{5}\\[2pt]72^{5}&=19^{5}+43^{5}+46^{5}+47^{5}+67^{5}\\[2pt]94^{5}&=21^{5}+23^{5}+37^{5}+79^{5}+84^{5}\\[2pt]107^{5}&=7^{5}+43^{5}+57^{5}+80^{5}+100^{5}\end{aligned}}}$$

(Lander & Parkin, 1966);^[10][11][12] (Lander, Parkin, Selfridge, smallest, 1967);^[8] (Lander, Parkin, Selfridge, second smallest, 1967);^[8] (Sastry, 1934, third smallest).^[8]

It has been known since 2002 that there are no solutions for k = 6 whose final term is ≤ 730000.^[13]

$${\displaystyle 568^{7}=127^{7}+258^{7}+266^{7}+413^{7}+430^{7}+439^{7}+525^{7}}$$

(M. Dodrill, 1999).^[14]

$${\displaystyle 1409^{8}=90^{8}+223^{8}+478^{8}+524^{8}+748^{8}+1088^{8}+1190^{8}+1324^{8}}$$

(S. Chase, 2000).^[15]

• 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

• Tito Piezas III, A Collection of Algebraic Identities Archived 2011-10-01 at the Wayback Machine
• 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)
• R. Gerbicz, 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!