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where m and k are restricted to the positive integers—that is, it is considered as a Diophantine equation. The only known solution is 11 + 21 = 31, and Paul Erdős conjectured that no further solutions exist. As of 2024, the problem remains open, but it has been shown that any further solutions must have m > 10109.[1]
Some care must be taken when comparing research papers on this equation, because some articles[2] study it in the form 1k + 2k + ⋯ + mk = (m + 1)k instead.
Constraints on solutions
In 1953, Leo Moser proved that, in any further solutions,[3]
p | (m − 1) implies (p − 1) | k,
m − 1 is squarefree,
k is even,
p | (m + 1) implies (p − 1) | k,
m + 1 is either squarefree or 4 times an odd squarefree number,
In 1999, Moser's method was refined to show that m > 1.485 × 109,321,155.[5]
In 2002, it was shown[6] that k must be a multiple of 23 · 3# · 5# · 7# · 19# · 1000#, where the symbol # indicates the primorial; that is, n# is the product of all prime numbers ≤ n. This number exceeds 5.7462 × 10427.
In 2009, it was shown that 2k / (2m – 3) must be a convergent of ln(2); in what the authors of that paper call “one of very few instances where a large scale computation of a numerical constant has an application”, it was then determined that m > 2.7139 × 101,667,658,416.[1]
(m2 – 1) (4m2 – 1) / 12 has at least 4,990,906 prime factors, none of which are repeated.
The number 4,990,906 arises from the fact that where pn is the nth prime number.
Moser's method
First, let p be a prime factor of m − 1. Leo Moser showed[3] that this implies that p − 1 divides k and that
(1)
which upon multiplying by p yields
This in turn implies that m − 1 must be squarefree. Furthermore, it is easily checked that, in any nontrivial solutions, m − 1 > 2; since all squarefree numbers in this range must have at least one odd prime factor, the fact that p − 1 divides k implies that k must be even.
One congruence of the form (1) exists for each prime factor p of m − 1. Multiplying all of them together yields
Expanding out the product yields
where the higher-order terms are products of multiple factors of the form (m − 1) / p, with different values of p in each factor. These terms are all divisible by m − 1, so they all drop out of the congruence, yielding
The congruences (2), (3), (4), and (5) are quite restrictive; for example, the only values of m < 1000 which satisfy (2) are 3, 7, and 43, and these are ruled out by (4).
We now split into two cases: either m + 1 is even, or it is odd.
In the case that m + 1 is even, adding the left-hand sides of the congruences (2), (3), (4), and (5) must yield an integer, and this integer must be at least 4. Furthermore, the Euclidean algorithm shows that no prime p > 3 can divide more than one of the numbers in the set {m − 1, m + 1, 2m − 1, 2m + 1}, and that 2 and 3 can divide at most two of these numbers. Letting M = (m − 1) (m + 1) (2m − 1) (2m + 1), we then have
(6)
Since there are no nontrivial solutions with m < 1000, the part of the LHS of (6) outside the sigma cannot exceed 0.006; we therefore have
Therefore, if , then . In Moser's original paper [3], bounds on the prime-counting function are used to observe that
Therefore, M must exceed the product of the first 10,000,000 primes. This in turn implies that m > 10106 in this case.
In the case that m + 1 is odd, we cannot use (3), so instead of (6) we obtain
where N = (m − 1) (2m − 1) (2m + 1). On the surface, this appears to be a weaker condition than (6), but since m + 1 is odd, the prime 2 cannot appear on the greater side of this inequality, and it turns out to be a stronger restriction on m than the other case.
Therefore any nontrivial solutions have m > 10106.
In 1999, this method was refined by using computers to replace the prime-counting estimates with exact computations; this yielded the bound m > 1.485 × 109,321,155.[5]
Bounding the ratio m / (k + 1)
Let Sk(m) = 1k + 2k + ⋯ + (m – 1)k. Then the Erdős–Moser equation becomes Sk(m) = mk.
Method 1: Integral comparisons
By comparing the sum Sk(m) to definite integrals of the function xk, one can obtain the bounds 1 < m / (k + 1) < 3.
The sum Sk(m) = 1k + 2k + ⋯ + (m – 1)k is the upper Riemann sum corresponding to the integral in which the interval has been partitioned on the integer values of x, so we have
By hypothesis, Sk(m) = mk, so
which leads to
(7)
Similarly, Sk(m) is the lower Riemann sum corresponding to the integral in which the interval has been partitioned on the integer values of x, so we have
Substituting this into (10) to eliminate mk yields
Reindexing the sum on the right with the substitution i = s + 1 yields
(12)
We already know from (11) that k + 1 < m. This leaves open the possibility that m = k + 2; however, substituting this into (12) yields
which is impossible for k > 1, since the sum contains only positive terms. Therefore any nontrivial solutions must have m ≠ k + 2; combining this with (11) yields
We therefore observe that the left-hand side of (12) is positive, so
(13)
Since k > 1, the sequence is decreasing. This and (13) together imply that its first term (the term with s = 1) must be positive: if it were not, then every term in the sum would be nonpositive, and therefore so would the sum itself. Thus,
^ abcKrzysztofek, Bronisław (1966). "O Rówaniu 1n + ... + mn = (m + 1)n·k"(PDF). Zeszyty Naukowe Wyżsej Szkoły Pedagogicznej w Katowicach—Sekcja Matematyki (in Polish). 5: 47–54. Archived from the original(PDF) on 2024-05-13. Retrieved 2024-05-13.