# Euler's sum of powers conjecture

Short description: Disproved conjecture in number theory

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

= bk
nk

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

= bk
, 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. 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 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. This was published in a paper comprising just two sentences. 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. His smallest counterexample was

26824404 + 153656394 + 187967604 = 206156734.

A particular case of Elkies' solutions can be reduced to the identity

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

where

u2 = 22030 + 28849v56158v2 + 36941v331790v4.

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

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.

## Generalizations

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

$\displaystyle{ \sum_{i=1}^{n} a_i^k = \sum_{j=1}^{m} b_j^k }$,

where aibj are positive integers for all 1 ≤ in and 1 ≤ jm, then m + nk. 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 nk − 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 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
$\displaystyle{ (3a^2+5ab-5b^2)^3 + (4a^2-4ab+6b^2)^3 + (5a^2-5ab-3b^2)^3 = (6a^2-4ab+4b^2)^3 . }$ 
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
$\displaystyle{ a^3(a^3+2b^3)^3 = a^3(a^3-b^3)^3+b^3(a^3-b^3)^3+b^3(2a^3+b^3)^3. }$
The number 2 100 0003 can be expressed as the sum of three cubes in nine different ways.

### k = 4

958004 + 2175194 + 4145604 = 4224814 (R. Frye, 1988)
304 + 1204 + 2724 + 3154 = 3534 (R. Norrie, 1911)

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

### k = 5

275 + 845 + 1105 + 1335 = 1445 (Lander & Parkin, 1966)
195 + 435 + 465 + 475 + 675 = 725 (Lander, Parkin, Selfridge, smallest, 1967)
215 + 235 + 375 + 795 + 845 = 945 (Lander, Parkin, Selfridge, second smallest, 1967)
75 + 435 + 575 + 805 + 1005 = 1075 (Sastry, 1934, third smallest)

### k = 6

As of 2002, there are no solutions for $\displaystyle{ k=6 }$ whose final term is ≤ 730000.

### k = 7

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

### k = 8

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