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 $k$th powers of positive integers is itself a $k$th power, then $n$ is greater than or equal to $k$:

$$a_1^k + a_2^k + \dots + a_n^k = b^k \implies n \ge k$$

The conjecture represents an attempt to generalize Fermat's Last Theorem, which is the special case $n = 2$: if $$a_1^k + a_2^k = b^k,$$ 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 $k ≥ 6$.

Background
Euler was aware of the equality 59$4$ + 158$4$ = 133$4$ + 134$4$ 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 3$3$ + 4$3$ + 5$3$ = 6$3$ or the taxicab number 1729. The general solution of the equation $$x_1^3+x_2^3=x_3^3+x_4^3$$ is

$$\begin{align} x_1 &=\lambda( 1-(a-3b)(a^2+3b^2)) \\[2pt] x_2 &=\lambda( (a+3b)(a^2+3b^2)-1 )\\[2pt] x_3 &=\lambda( (a+3b)-(a^2+3b^2)^2 )\\[2pt] x_4 &= \lambda( (a^2+3b^2)^2-(a-3b)) \end{align}$$

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

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: $$\begin{align} 144^5 &= 27^5 + 84^5 + 110^5 + 133^5 \\ 14132^5 &= (-220)^5 + 5027^5 + 6237^5 + 14068^5 \\  85359^5 &= 55^5 + 3183^5 + 28969^5 + 85282^5 \end{align}$$ (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. His smallest counterexample was $$20615673^4 = 2682440^4 + 15365639^4 + 18796760^4.$$

A particular case of Elkies' solutions can be reduced to the identity $$(85v^2 + 484v - 313)^4 + (68v^2 - 586v + 10)^4 + (2u)^4 = (357v^2 - 204v + 363)^4,$$ where $$u^2 = 22030 + 28849v - 56158v^2 + 36941v^3 - 31790v^4.$$ This is an elliptic curve with a rational point at $v_{1} = −31⁄467$. From this initial rational point, one can compute an infinite collection of others. Substituting $v_{1}$ into the identity and removing common factors gives the numerical example cited above.

In 1988, Roger Frye found the smallest possible counterexample $$95800^4 + 217519^4 + 414560^4 = 422481^4$$ 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.

Generalizations


In 1967, L. J. Lander, T. R. Parkin, and John Selfridge conjectured that if
 * $$\sum_{i=1}^{n} a_i^k = \sum_{j=1}^{m} b_j^k$$,

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
 * $$\sum_{i=1}^{n} a_i^k = b^k$$

(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 for more data.

$k = 3$

 * 3$3$ + 4$3$ + 5$3$ = 6$3$ (Plato's number 216)


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

$$(3a^2+5ab-5b^2)^3 + (4a^2-4ab+6b^2)^3 + (5a^2-5ab-3b^2)^3 = (6a^2-4ab+4b^2)^3$$


 * A cube as the sum of three cubes can also be parameterized in one of two ways:

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


 * The number 2 100 0003 can be expressed as the sum of three cubes in nine different ways.

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

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

(Lander & Parkin, 1966); (Lander, Parkin, Selfridge, smallest, 1967); (Lander, Parkin, Selfridge, second smallest, 1967); (Sastry, 1934, third smallest).

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

$k = 7$
$$568^7 = 127^7 + 258^7 + 266^7 + 413^7 + 430^7 + 439^7 + 525^7$$

(M. Dodrill, 1999).

$k = 8$
$$1409^8 = 90^8 + 223^8 + 478^8 + 524^8 + 748^8 + 1088^8 + 1190^8 + 1324^8 $$

(S. Chase, 2000).