User:LucasBrown/EMEq

This is a development page for Erdős–Moser equation.

Unsolved problem in mathematics:

Does the Erdős–Moser equation have solutions other than 1¹+2¹=3¹?

(more unsolved problems in mathematics)

In number theory, the Erdős–Moser equation is

$1^{k}+2^{k}+\cdots +(m-1)^{k}=m^{k},$

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.

Some care must be taken when comparing research papers on this equation, because some articles examining the problem study it in the form 1^k + 2^k + ⋯ + m^k = (m + 1)^k instead.

Constraints on solutions

In 1953, Leo Moser proved that, in any further solutions, 2 must divide k and that m ≥ 10^1,000,000.

In 1966, it was shown that

1. m – 1 must be squarefree,
2. p – 1 divides k for every prime p dividing m – 1,
3. 6 ≤ k + 2 < m < 3 (k + 1) / 2,
4. k must be even,
5. m – 1 cannot be prime, and
6. $\sum _{p|m-1}{\frac {m-1}{p}}\equiv -1{\pmod {m-1}}.$

In 1994, it was shown that

1. lcm(1,2,…,200) divides k,
2. ord₂(m − 3) = ord₂(k) + 3, where ord₂(n) is the 2-adic valuation of n,
3. for any prime p divding m, we have k ≢ 0, 2 (mod p − 1),
4. m ≡ 3 (mod 8),
5. and that any prime factor of m must be irregular and > 10000.

In 1999, Moser's method was refined to show that m > 1.485 × 10^9,321,155.

In 2002, it was shown 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 the very few instances where a large scale computation of a numerical constant has an application”, it was then determined that m > 2.7139 × 10^{1,667,658,416}.

In 2010, it was shown that

1. k is even,
2. m ≡ 3 (mod 8) and m ≡ ±1 (mod 3),
3. m – 1, (m + 1) / 2, 2m – 1, and 2m + 1 are all squarefree,
4. if the prime number p divides at least one of the numbers from the previous line, then p – 1 divides k, and
5. (m² – 1) (4m² – 1) / 12 has at least 4,990,906 prime factors, none of which are repeated.

This number 4,990,906 arises from the fact that $\sum _{n=1}^{4990905}{\frac {1}{p_{n}}}<{\frac {19}{6}}<\sum _{i=1}^{4990906}{\frac {1}{p_{n}}},$ where p_n is the n^th prime number.

Moser's method

TODO

Bounding the ratio m / (k + 1)

Let S_k(m) = 1^k + 2^k + ⋯ + (m – 1)^k. Then the Erdős–Moser equation becomes S_k(m) = m^k.

The sum S_k(m) = 1^k + 2^k + ⋯ + (m – 1)^k is the upper Riemann sum corresponding to the integral ${\textstyle \int _{0}^{m-1}x^{k}\,\mathrm {d} x}$ in which the interval has been partitioned on the integer values of the integration variable, so we have

$S_{k}(m)>\int _{0}^{m-1}x^{k}\;\mathrm {d} x.$

By hypothesis, S_k(m) = m^k, so

$m^{k}>{\frac {(m-1)^{k+1}}{k+1}},$

which leads to

${\frac {m}{k+1}}<\left(1+{\frac {1}{m-1}}\right)^{k+1}.$

(1)

Similarly, S_k(m) is the lower Riemann sum corresponding to the integral ${\textstyle \int _{1}^{m}x^{k}\,\mathrm {d} x}$, so we have

$S_{k}(m)\leq \int _{1}^{m}x^{k}\;\mathrm {d} x.$

By hypothesis, S_k(m) = m^k, so

$m^{k}\leq {\frac {m^{k+1}-1}{k+1}}<{\frac {m^{k+1}}{k+1}},$

and so

$k+1<m.$

(2)

Applying this to (1) yields

${\frac {m}{k+1}}<\left(1+{\frac {1}{m-1}}\right)^{m}=\left(1+{\frac {1}{m-1}}\right)^{m-1}\cdot \left({\frac {m}{m-1}}\right)<e\cdot {\frac {m}{m-1}}.$

Computation shows that there are no nontrivial solutions with m ≤ 10, so we have

${\frac {m}{k+1}}<e\cdot {\frac {11}{11-1}}<3.$

Combining this with (2) yields 1 < m / (k + 1) < 3. A more elaborate analysis along these lines brings the upper bound down to 3/2.

Continued fractions

By expanding the Taylor series of (1 − y)^k e^ky about y = 0, one finds

$(1-y)^{k}=e^{-ky}\left(1-{\frac {k}{2}}y^{2}-{\frac {k}{3}}y^{3}+{\frac {k(k-2)}{8}}y^{4}+{\frac {k(5k-6)}{30}}y^{5}+O(y^{6})\right)\quad {\text{as }}y\rightarrow 0.$

More elaborate analysis sharpens this to

${\begin{aligned}e^{-ky}\left(1-{\frac {k}{2}}y^{2}-{\frac {k}{3}}y^{3}+{\frac {k(k-2)}{8}}y^{4}+{\frac {k(5k-6)}{30}}y^{5}-{\frac {k^{3}}{6}}y^{6}\right)\\<(1-y)^{k}<e^{-ky}\left(1-{\frac {k}{2}}y^{2}-{\frac {k}{3}}y^{3}+{\frac {k(k-2)}{8}}y^{4}+{\frac {k^{2}}{2}}y^{5}\right)\end{aligned}}$ (3)

for k > 8 and 0 < y < 1.

The Erdős–Moser equation is equivalent to

$1=\sum _{j=1}^{m-1}\left(1-{\frac {j}{m}}\right)^{k}.$

Applying (3) to each term in this sum yields

${\begin{aligned}S_{0}-{\frac {k}{2m^{2}}}S_{2}-{\frac {k}{3m^{3}}}S_{3}+{\frac {k(k-2)}{8m^{4}}}S_{4}+{\frac {k(5k-6)}{30m^{5}}}S_{5}-{\frac {k^{3}}{6m^{6}}}S_{6}\\<\sum _{j=1}^{m-1}\left(1-{\frac {j}{m}}\right)^{k}<S_{0}-{\frac {k}{2m^{2}}}S_{2}-{\frac {k}{3m^{3}}}S_{3}+{\frac {k(k-2)}{8m^{4}}}S_{4}+{\frac {k^{2}}{2m^{5}}}S_{5},\end{aligned}}$

where $S_{n}=\sum _{j=1}^{m-1}j^{n}z^{j}$ and z = e^−k/m.

By , we have e⁻¹ < z < e^−1/2; therefore 1 / (1 − z) < 1 / (1 − e^−1/2) < 3...

TODO

Further manipulation eventually yields

$\left\vert 1-{\frac {z}{1-z}}+{\frac {k}{2m^{2}}}{\frac {z^{2}+z}{(1-z)^{3}}}+{\frac {k}{3m^{3}}}{\frac {z^{3}+4z^{2}+z}{(1-z)^{4}}}-{\frac {k(k-2)}{8m^{4}}}{\frac {z^{4}+11z^{3}+11z^{2}+z}{(1-z)^{5}}}\right\vert <{\frac {110000}{m^{3}}}.$ (4)

We already know that k/m is bounded as m → ∞; making the ansatz k/m = c + O(1/m), and therefore z = e^−c + O(1/m), and substituting it into (4) yields

$1-{\frac {1}{e^{c}-1}}=O\left({\frac {1}{m}}\right)\quad {\text{as }}m\rightarrow \infty ;$

therefore c = ln(2). We therefore have

${\frac {k}{m}}=\ln(2)+{\frac {a}{m}}+{\frac {b}{m^{2}}}+O\left({\frac {1}{m^{3}}}\right)\quad {\text{as }}m\rightarrow \infty ,$ (5)

and so

${\frac {1}{z}}=e^{k/m}=2+{\frac {2a}{m}}+{\frac {a^{2}+2b}{m^{2}}}+O\left({\frac {1}{m^{3}}}\right)\quad {\text{as }}m\rightarrow \infty .$

Substituting these formulas into (4) yields a = −3 ln(2) / 2 and b = (3 ln(2) − 25/12) ln(2). Putting these into (5) and setting c₁ = 25/12 − 3 ln(2) yields

${\frac {k}{m}}=\ln(2)\left(1-{\frac {3}{2m}}-{\frac {c_{1}}{m^{2}}}+O\left({\frac {1}{m^{3}}}\right)\right)\quad {\text{as }}m\rightarrow \infty .$

TODO: recall that z here is e^(-k/m), but z gets reciprocated in the article

This proceeds by deriving the inequality

$0<\ln(2)-{\frac {2k}{2m-3}}<{\frac {0.0111}{(2m-3)^{2}}}$

for m > 10⁹, recalling that Moser showed that indeed m > 10⁹, and then invoking Legendre's theorem on continued fractions to conclude that 2k / (2m – 3) must be a convergent to ln(2).

TODO: Explain the derivation of that inequality.

General TODOs

TODO: Find links to and .

TODO: What level of thesis is for?

TODO: Find a way to cite .

Carlitz's article on Staudt-Clausen: