Euler's sum of powers conjecture

Euler's sum of powers conjecture is a statement in number theory, proposed by Leonhard Euler in 1769, asserting that for any integer k \geq 3, no k-th power of a positive integer can be expressed as a sum of fewer than k positive k-th powers of positive integers.^[1] This conjecture generalized Fermat's Last Theorem, which corresponds to the case of two summands being impossible for k > 2, by extending the idea to arbitrary numbers of summands less than k.^[2] The hypothesis stood unchallenged for nearly two centuries, reflecting Euler's attempt to describe the minimal number of terms required in such Diophantine equations.^[3] The conjecture held for k = 3, where sums of two positive cubes cannot equal a cube—a result aligned with Fermat's theorem—and non-trivial solutions with exactly three cubes do exist, such as $3^3 + 4^3 + 5^3 = 6^3.^[2] However, it was first disproven in 1966 for k = 5 by L. J. Lander and T. R. Parkin, who discovered that four fifth powers can sum to a fifth power: $27^5 + 84^5 + 110^5 + 133^5 = 144^5.^[1] This counterexample, found via computer search on a CDC 6600, marked one of the shortest publications in mathematical history, consisting primarily of the equation itself.^[2] A counterexample for k = 4 followed in 1988, when Noam D. Elkies constructed an infinite family of solutions showing that three fourth powers can sum to a fourth power, with one explicit instance being $2682440^4 + 15365639^4 + 18796760^4 = 20615673^4.^[4] Elkies's method relied on elliptic curves to generate parametric solutions, disproving the conjecture for this case as well.^[2] Shortly thereafter, Roger Frye used extensive computer search to identify the smallest such solution in positive integers: $95800^4 + 217519^4 + 414560^4 = 422481^4.^[3] These discoveries highlighted the power of computational methods in number theory and prompted further research into the generalized Waring's problem for equal powers. For higher k \geq 6, no counterexamples with fewer than k summands have been found, but it is widely believed that the minimal number g(k) of positive k-th powers needed to sum to a k-th power satisfies g(k) \leq k-1, with current bounds indicating g(k) \geq 4 in general.^[3] The conjecture's disproof has influenced studies in additive number theory, including equal sums of like powers and the search for parametric families of solutions.^[2]

Background

Historical Origins

Leonhard Euler proposed the sum of powers conjecture in 1769 while engaged in his extensive studies of Diophantine equations and number theory.^[3] This proposal emerged from his broader investigations into the properties of powers and their sums, as documented in historical accounts of 18th-century mathematics.^[1] Euler's motivation was rooted in generalizing earlier observations about equal sums of powers, particularly building on Pierre de Fermat's work regarding the impossibility of certain power equalities for exponents greater than 2.^[2] Fermat had demonstrated that no two positive integers raised to the fourth power sum to another fourth power, and Euler extended this line of inquiry to consider the minimal number of terms required for such sums across higher exponents. By the end of the 18th century, the conjecture had become a cornerstone reference in Diophantine analysis, though it remained unproven and subject to further scrutiny.^[1]

Formal Statement

Euler's sum of powers conjecture, proposed by Leonhard Euler in 1769, asserts that for every integer k > 2, no positive integer b^k can be expressed as a sum of fewer than k positive k-th powers of positive integers strictly less than b. Formally, there do not exist positive integers a_1, \dots, a_n, b with n < k and $1 \leq a_i < b for all i such that

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

This formulation assumes positive integers only, thereby excluding zero and the trivial identity b^k = b^k (corresponding to n=1 and a_1 = b). The conjecture generalizes Fermat's Last Theorem, which corresponds to the case n=1 (impossible for k \geq 3), by extending the prohibition to sums with up to k-1 terms. It posits that at least k positive k-th powers are required to sum to any k-th power.^[3]

Results for Small Exponents

Exponent 2: Squares

For the exponent k=2, Euler's sum of powers conjecture posits that at least two positive squares are required to sum to another square b^2. This holds because a single positive square a^2 with $1 \leq a < b cannot equal b^2, as it would imply a = b, violating the strict inequality inherent in non-trivial representations. However, sums of exactly two positive squares frequently equal a square, as demonstrated by Pythagorean triples, such as the primitive example $3^2 + 4^2 = 5^2. All primitive Pythagorean triples (a, b, c) with a^2 + b^2 = c^2 can be generated using integers m > n > 0 that are coprime and of opposite parity via the formulas a = m^2 - n^2, b = 2mn, c = m^2 + n^2; for instance, m=2, n=1 yields the triple (3, 4, 5).^[5] This phenomenon ties into the theory of sums of two squares, which characterizes numbers expressible as x^2 + y^2 for integers x, y \geq 0: a positive integer has such a representation if and only if every prime congruent to 3 modulo 4 in its prime factorization has even exponent. Since b^2 is a perfect square, all exponents in its factorization are even, satisfying the condition and allowing b^2 = x^2 + y^2 (possibly with one term zero). For positive x, y > 0, the representations correspond precisely to Pythagorean triples, confirming that two terms suffice when possible.^[6] Lagrange's four-square theorem further contextualizes this by proving that every natural number, including any square b^2, can be written as the sum of at most four integer squares. While this guarantees a representation with four terms, the minimal number for non-trivial positive sums to a square is two, as provided by Pythagorean triples, aligning directly with the conjecture's requirement for k=2. No counterexamples exist, as single-term sums are impossible non-trivially, and two-term sums are abundantly realized.^[7]

Exponent 3: Cubes

For the exponent k=3, Euler's conjecture asserts that at least three positive cubes are needed to sum to another cube, excluding the trivial identity of a single cube equaling itself. The single-term case is excluded as trivial, while sums of two positive cubes cannot equal a cube, a result established by Euler's proof of Fermat's Last Theorem for exponent 3 using the method of infinite descent, published in 1770.^[8] This impossibility implies that the minimal number g(3)=3, confirming the conjecture holds for cubes.^[3] The smallest non-trivial example is $3^3 + 4^3 + 5^3 = 6^3, or $27 + 64 + 125 = 216, discovered by Euler during his investigations into sums of powers.^[3] This identity illustrates that three cubes can indeed sum to a fourth cube, aligning with the conjectured minimal number. Infinite families of such identities exist, providing parametric solutions for positive integers a, b > 0. One such family, originally given by Ramanujan, is

\begin{aligned} &(3a^2 + 5ab - 5b^2)^3 + (4a^2 - 4ab + 6b^2)^3 + (5a^2 - 5ab - 3b^2)^3 \\ &= (6a^2 - 4ab + 4b^2)^3. \end{aligned}

For a=1, b=0, this reduces to the smallest example above (up to scaling). Substituting positive integers for a and b yields infinitely many distinct solutions, demonstrating the abundance of three-cube representations of cubes.^[9]

Counterexamples

Exponent 5: Initial Disproof

In 1966, L. J. Lander and T. R. Parkin provided the first counterexample to Euler's sum of powers conjecture for exponent 5, demonstrating that four fifth powers can sum to another fifth power, contrary to the expectation of requiring at least five terms.^[1] Their discovery, communicated by J. D. Swift and published in the Bulletin of the American Mathematical Society, marked a significant breakthrough in the study of Diophantine equations involving equal powers.^[10] The counterexample is given by the equation

$27^5 + 84^5 + 110^5 + 133^5 = 144^5,

which Lander and Parkin identified through a direct computational search on a CDC 6600 computer.^[1] This search systematically checked combinations of positive integers, revealing the equality as the smallest such instance where four fifth powers sum to a fifth power.^[10] Both sides of the equation evaluate to 61,917,364,224, confirming the numerical balance.^[11] This result establishes that the minimal number g(5) of fifth powers required to sum to another fifth power satisfies g(5) \leq 4, directly disproving the conjecture for k=5.^[1] The finding, one of the shortest papers in the journal's history, underscored the power of early computational methods in number theory and prompted further investigations into the conjecture for other exponents.^[10]

Exponent 4: Confirmed Disproof

In 1988, Noam Elkies constructed an infinite family of counterexamples for exponent 4 using a parametric method based on elliptic curves, providing the first disproof for this case with the explicit example $2682440^4 + 15365639^4 + 18796760^4 = 20615673^4.^[2] This demonstrated that three positive fourth powers can sum to another fourth power, confirming that the minimal number g(4) \leq 3, which is less than the conjectured 4. Later in 1988, Roger Frye, inspired by Elkies' work, conducted an exhaustive computational search using optimized algorithms on a Sun workstation over 100 days, identifying the smallest known primitive solution in positive integers:
$95800^4 + 217519^4 + 414560^4 = 422481^4.
Frye's search systematically checked combinations up to bases near 1 million, verifying no smaller solutions exist. This remains the smallest solution in terms of the largest base as of 2025.^[3] These discoveries highlighted the synergy of theoretical and computational approaches in resolving the case for even powers.

Status for Higher Exponents

Exponent 6: Computational Searches

No counterexample to Euler's conjecture is known for the exponent k=6 using fewer than 6 positive sixth powers summing to another sixth power, indicating that the minimal number g(6) of sixth powers needed to represent a sixth power is at least 6, although this has not been proven.^[3] Trivial solutions with exactly 6 terms exist through parametric identities, such as those derived from algebraic constructions, but the interest lies in non-trivial representations with m < 6 terms. Extensive computational efforts to identify a counterexample with m=5 have been unsuccessful. No counterexamples have been verified as of 2025.^[3] While the minimal number remains unresolved, results from the Waring problem provide an upper bound: every positive integer, including every sixth power, can be expressed as the sum of at most 73 sixth powers, with the more refined general Waring number indicating that 23 sixth powers suffice for all sufficiently large integers and thus for large sixth powers.^[12] These bounds establish that representations with m \leq 23 always exist, narrowing the focus to whether fewer than 6 terms are possible for the specific case of summing to a sixth power.

Exponents 7 and 8: Bounds and Open Questions

For the exponent k=7, no counterexamples have been found for representations of a seventh power as a sum of m<7 positive seventh powers. Extensive computational searches have not revealed solutions for m=6, and the conjecture remains open as of 2025.^[3] The case for k=8 is similarly unresolved, with exhaustive computational checks for small bases confirming no counterexamples involving m<8. However, it is possible that g(8)<8, where g(k) denotes the minimal m for which there exists a non-trivial solution to the equation with m terms.^[3] Probabilistic methods establish that g(k) ≤ k for sufficiently large k, though the precise value remains unknown for small k beyond 5.^[3] A central open question is whether g(k)=k-1 or smaller for k≥6; unlike the situation for k=4, no parametric families of solutions are known for higher exponents.^[3] It is widely believed that counterexamples exist for all k ≥ 4, though none have been found for k ≥ 6.

Lander-Parkin-Selfridge Conjecture

The Lander-Parkin-Selfridge conjecture, proposed by L. J. Lander, T. R. Parkin, and J. L. Selfridge in 1967, states that if \sum_{i=1}^m a_i^k = \sum_{j=1}^n b_j^k where the a_i and b_j are distinct positive integers, then m + n \geq k + 1. This conjecture was motivated by the counterexample to Euler's sum of powers conjecture for k=5 presented in related work by the authors, which highlighted the need for a refined lower bound on the total number of terms in such equalities; it holds for small k but remains unproven in general. The general form of the key equation is \sum_{i=1}^m a_i^k = \sum_{j=1}^n b_j^k, implying m + n \geq k + 1 under the distinctness condition. For verification, consider k=3, where the minimal equal sums require at least 4 terms total; a representative example is the taxicab number $1729 = 1^3 + 12^3 = 9^3 + 10^3, with m=2, n=2. The conjecture has been verified computationally for small k, with no counterexamples found, but it remains open in general.^[13]

Multigrade Identities and Extensions

Multigrade identities generalize the idea of equal sums of like powers by requiring the equality to hold simultaneously for multiple consecutive exponents. A (k, l)-multigrade equation is defined as \sum_{i=1}^l n_i^j = \sum_{i=1}^l m_i^j for j = 1, 2, \dots, k, where the m_i and n_i are distinct positive integers forming two sets of size l.^[14] These identities can be normalized by adding a constant to all terms while preserving the equalities, often starting with the smallest element as 1.^[14] A simple example is the (2,3)-multigrade with sets \{1, 6, 8\} and \{2, 4, 9\}, where the sums equal 15 for j=1 and 101 for j=2:

$1 + 6 + 8 = 2 + 4 + 9 = 15,

$1^2 + 6^2 + 8^2 = 2^2 + 4^2 + 9^2 = 101.

^[14] For higher degrees, a (4,6)-multigrade uses sets \{1, 5, 8, 12, 18, 19\} and \{2, 3, 9, 13, 16, 20\}, with equal sums up to j=4: 63 for j=1, 919 for j=2, 15057 for j=3, and 260755 for j=4.^[14] Such identities have been constructed using binary quadratic forms for exponents including the third, fourth, fifth, and seventh powers.^[14] The Prouhet–Tarry–Escott problem provides a framework for these identities, seeking distinct sets of n integers with equal power sums for the first m exponents, and serves as a special case of multigrades.^[15] Computational searches have yielded solutions for m up to 11, such as Chen's 1999 construction for sets of size 12 equal through the 11th power.^[15] These efforts draw inspiration from related problems like the Lander–Parkin–Selfridge conjecture on minimal terms for equal sums at a single exponent.^[15] Ramanujan developed parametric identities yielding infinite families of solutions for equal sums of like powers at specific exponents. For cubes, given a^2 + a\beta + \beta^2 = 3\lambda y^2, the identity (a + \lambda^2 y)^3 + (\lambda \beta + y)^3 = (\lambda a + y)^3 + (1 + \lambda^2 y)^3 holds; setting a=3, \beta=0, y=1, \lambda=3 gives the famous $1^3 + 12^3 = 9^3 + 10^3 = 1729.^[16] For fourth powers, a parametric form expresses one fourth power as a sum of five fourth powers, such as $4^4 + 6^4 + 8^4 + 9^4 + 14^4 = 15^4 when parameters s=1, t=0.^[16] These connect to broader themes in modular forms through Ramanujan's work on power sums.^[16] Extensions to mixed exponents consider whether sums of m kth powers can equal sums of n lth powers for k \neq l, with questions on minimalities analogous to Euler's original conjecture. Computational examples exist up to k=10, often involving signed integers for balance, such as a (9,10)-multigrade with alternating signs yielding zero sums for odd exponents up to 9.^[14] For positive integers, open questions persist on whether Euler-like minimal term counts hold across differing exponents.^[14]