Erdős–Moser equation

{{unsolved|mathematics|2=Does the Erdős–Moser equation have solutions other than {{math|1=1¹+2¹=3¹}}?}}

In number theory, the Erdős–Moser equation is

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

where {{mvar|m}} and {{mvar|k}} are restricted to the positive integers—that is, it is considered as a Diophantine equation. The only known solution is {{math|1=1¹ + 2¹ = 3¹}}, and Paul Erdős conjectured that no further solutions exist. Any further solutions must have {{math|1=m > 10^10⁹}}.

Throughout this article, {{mvar|p}} refers exclusively to prime numbers.

Constraints on solutions

In 1953, Leo Moser proved that, in any further solutions,

{{mvar|k}} is even,
{{math|1=p | (m − 1)}} implies {{math|1=(p − 1) | k}},
{{math|1=p | (m + 1)}} implies {{math|1=(p − 1) | k}},
{{math|1=p | (2m + 1)}} implies {{math|1=(p − 1) | k}},
{{math|1=m − 1}} is squarefree,
{{math|1=m + 1}} is either squarefree or 4 times an odd squarefree number,
{{math|1=2m − 1}} is squarefree,
{{math|1=2m + 1}} is squarefree,
$\sum_{p|(m-1)} \frac{m-1}{p} \equiv -1 \pmod{m-1},$
$\sum_{p|(m+1)} \frac{m+1}{p} \equiv -2 \pmod{m+1} \quad \text{(if }m+1\text{ is even)},$
$\sum_{p|(2m-1)} \frac{2m-1}{p} \equiv -2 \pmod{2m-1},$
$\sum_{p|(2m+1)} \frac{2m+1}{p} \equiv -4 \pmod{2m+1},$ and
{{math|m > 10^10⁶}}.

In 1966, it was additionally shown that

{{math|1=6 ≤ k + 2 < m < 3 (k + 1) / 2}}, and
{{math|1=m – 1}} cannot be prime.

In 1994, it was shown that

Least common multiple divides {{mvar|k}},
{{math|1=m ≡ 3 (mod 2^{ord₂(k) + 3})}}, where {{math|1=ord₂(n)}} is the 2-adic valuation of {{mvar|n}}; equivalently, {{math|1=ord₂(m − 3) = ord₂(k) + 3}},
for any odd prime {{Mvar|p}} divding {{Mvar|m}}, we have {{Math|k ≢ 0, 2 (mod p − 1)}},
any prime factor of {{mvar|m}} must be irregular and > 10000.

In 1999, Moser's method was refined to show that {{math|1=m > 1.485 × 10^9,321,155}}.

In 2002, it was shown{{r|kellner2002|at=§4}} that {{mvar|k}} must be a multiple of {{math|1=2³ · 3# · 5# · 7# · 19# · 1000#}}, where the symbol {{mvar|#}} indicates the primorial; that is, {{math|1=n#}} is the product of all prime numbers {{math|≤ n}}. This number exceeds {{math|1=5.7462 × 10⁴²⁷}}.

In 2009, it was shown that {{math|1=2k / (2m – 3)}} must be a convergent of Natural logarithm of 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 {{math|1=m > 2.7139 × 10^{1,667,658,416}}}.

In 2010, it was shown that

{{math|1=m ≡ 3 (mod 8)}} and {{math|1=m ≡ ±1 (mod 3)}}, and
{{math|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 {{math|1={{resize|150%|∑}}{{su|p=4990905|b=n=1}} {{sfrac|p{{sub|n}}}} < {{sfrac|19|6}} < {{resize|150%|∑}}{{su|p=4990906|b=n=1}} {{sfrac|p{{sub|n}}}},}} where {{math|1=p_n}} is the {{mvar|n}}^th prime number.

Moser's method

First, let {{mvar|p}} be a prime factor of {{math|1=m − 1}}. Leo Moser showed that this implies that {{math|1=p − 1}} divides {{mvar|k}} and that

{{NumBlk|:| $1 + \frac{m - 1}{p} \equiv 0 \pmod{p},$ | {{EquationRef|1}} | LnSty=0.37ex dotted Gainsboro}}

which upon multiplying by {{mvar|p}} yields

: $p + m - 1 \equiv 0 \pmod{p^2}.$

This in turn implies that {{math|m − 1}} must be squarefree. Furthermore, since nontrivial solutions have {{math|1=m − 1 > 2}} and since all squarefree numbers in this range must have at least one odd prime factor, the assumption that {{math|1=p − 1}} divides {{mvar|k}} implies that {{mvar|k}} must be even.

One congruence of the form ({{EquationNote|1}}) exists for each prime factor {{mvar|p}} of {{math|1=m − 1}}. Multiplying all of them together yields

: $\prod_{p|(m-1)} \left( 1 + \frac{m-1}{p} \right) \equiv 0 \pmod{m-1}.$

Expanding out the product yields

: $1 + \sum_{p|(m-1)} \frac{m-1}{p} + (\text{higher-order terms}) \equiv 0 \pmod{m-1},$

where the higher-order terms are products of multiple factors of the form {{math|1=(m − 1) / p}}, with different values of {{mvar|p}} in each factor. These terms are all divisible by {{math|1=m − 1}}, so they all drop out of the congruence, yielding

: $1 + \sum_{p|(m-1)} \frac{m-1}{p} \equiv 0 \pmod{m-1}.$

Dividing out the modulus yields

{{NumBlk|:| $\frac{1}{m-1} + \sum_{p|(m-1)} \frac{1}{p} \equiv 0 \pmod{1}.$ | {{EquationRef|2}} | LnSty=0.37ex dotted Gainsboro}}

Similar reasoning yields the congruences

{{NumBlk|:| $\frac{2}{m+1} + \sum_{p|(m+1)} \frac{1}{p} \equiv 0 \pmod{1} \quad \text{(if }m+1\text{ is even)},$ | {{EquationRef|3}} | LnSty=0.37ex dotted Gainsboro}}

{{NumBlk|:| $\frac{2}{2m-1} + \sum_{p|(2m-1)} \frac{1}{p} \equiv 0 \pmod{1}, \quad \text{and}$ | {{EquationRef|4}} | LnSty=0.37ex dotted Gainsboro}}

{{NumBlk|:| $\frac{4}{2m+1} + \sum_{p|(2m+1)} \frac{1}{p} \equiv 0 \pmod{1}.$ | {{EquationRef|5}} | LnSty=0.37ex dotted Gainsboro}}

The congruences ({{EquationNote|2}}), ({{EquationNote|3}}), ({{EquationNote|4}}), and ({{EquationNote|5}}) are quite restrictive; for example, the only values of {{math|1=m < 1000}} which satisfy ({{EquationNote|2}}) are 3, 7, and 43, and these are ruled out by ({{EquationNote|4}}).

We now split into two cases: either {{math|1=m + 1}} is even, or it is odd.

In the case that {{math|1=m + 1}} is even, adding the left-hand sides of the congruences ({{EquationNote|2}}), ({{EquationNote|3}}), ({{EquationNote|4}}), and ({{EquationNote|5}}) must yield an integer, and this integer must be at least 4. Furthermore, the Euclidean algorithm shows that no prime {{math|1=p > 3}} can divide more than one of the numbers in the set {{math|1={{mset|m − 1, m + 1, 2m − 1, 2m + 1}}}}, and that 2 and 3 can divide at most two of these numbers. Letting {{math|1=M = (m − 1) (m + 1) (2m − 1) (2m + 1)}}, we then have

{{NumBlk|:| $\frac{1}{m-1} + \frac{2}{m+1} + \frac{2}{2m-1} + \frac{4}{2m+1} + \sum_{p|M} \frac{1}{p} \geq 4 - \frac{1}{2} - \frac{1}{3} = 3.1666....$ | {{EquationRef|6}} | LnSty=0.37ex dotted Gainsboro}}

Since there are no nontrivial solutions with {{math|1=m < 1000}}, the part of the LHS of ({{EquationNote|6}}) outside the sigma cannot exceed {{math|1=0.006}}; we therefore have

: $\sum_{p|M} \frac{1}{p} > 3.16.$

Therefore, if $\sum_{p \leq x} \frac{1}{p} < 3.16$ , then $M > \prod_{p \leq x} p$ . In Moser's original paper, bounds on the prime-counting function are used to observe that

: $\sum_{p \leq 10^7} \frac{1}{p} < 3.16.$

Therefore, {{mvar|M}} must exceed the product of the first 10,000,000 primes. This in turn implies that {{math|1=m > 10^10⁶}} in this case.

In the case that {{math|1=m + 1}} is odd, we cannot use ({{EquationNote|3}}), so instead of ({{EquationNote|6}}) we obtain

: $\frac{1}{m-1} + \frac{2}{2m-1} + \frac{4}{2m+1} + \sum_{p|N} \frac{1}{p} \geq 3 - \frac{1}{3} = 2.666...,$

where {{math|1=N = (m − 1) (2m − 1) (2m + 1)}}. On the surface, this appears to be a weaker condition than ({{EquationNote|6}}), but since {{math|1=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 {{mvar|m}} than the other case.

Therefore any nontrivial solutions have {{math|1=m > 10^10⁶}}.

In 1999, this method was refined by using computers to replace the prime-counting estimates with exact computations; this yielded the bound {{math|1=m > 1.485 × 10^9,321,155}}.{{r|butske1999|at=Thm 2}}

Bounding the ratio {{math|1=''m'' / (''k'' + 1)}}

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

= Method 1: Integral comparisons =

By comparing the sum {{math|1=S_k(m)}} to definite integrals of the function {{math|1=x^k}}, one can obtain the bounds {{math|1=1 < m / (k + 1) < 3}}.{{r|gallot2010|at=§1¶2}}

The sum {{math|1=S_k(m) = 1^k + 2^k + ⋯ + (m – 1)^k}} is the upper Riemann sum corresponding to the integral $\int_0^{m-1} x^k \, \mathrm{d}x$ in which the interval has been partitioned on the integer values of {{mvar|x}}, so we have

: $S_k(m) > \int_0^{m-1} x^k \; \mathrm{d}x.$

By hypothesis, {{math|1=S_k(m) = m^k}}, so

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

which leads to

{{NumBlk|:| $\frac{m}{k+1} < \left( 1 + \frac{1}{m-1} \right)^{k+1}.$ |{{EquationRef|7}}|LnSty=0.37ex dotted Gainsboro}}

Similarly, {{math|1=S_k(m)}} is the lower Riemann sum corresponding to the integral $\int_1^m x^k \, \mathrm{d}x$ in which the interval has been partitioned on the integer values of {{mvar|x}}, so we have

: $S_k(m) \leq \int_1^m x^k \; \mathrm{d}x.$

By hypothesis, {{math|1=S_k(m) = m^k}}, so

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

and so

{{NumBlk|:| $k+1 < m.$ |{{EquationRef|8}}|LnSty=0.37ex dotted Gainsboro}}

Applying this to ({{EquationNote|7}}) 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 {{math|1=m ≤ 10}}, so we have

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

Combining this with ({{EquationNote|8}}) yields {{math|1=1 < m / (k + 1) < 3}}, as desired.

= Method 2: Algebraic manipulations =

The upper bound {{math|1=m / (k + 1) < 3}} can be reduced to {{math|1=m / (k + 1) < 3/2}} using an algebraic method:{{r|krzysztofek1966|at=Lemat 4}}

Let {{mvar|r}} be a positive integer. Then the binomial theorem yields

: $(\ell + 1)^{r+1} = \sum_{i=0}^{r+1} \binom{r+1}{i} \ell^{r+1-i}.$

Summing over {{mvar|ℓ}} yields

: $\sum_{\ell=1}^{m-1} (\ell + 1)^{r+1} = \sum_{\ell=1}^{m-1} \left( \sum_{i=0}^{r+1} \binom{r+1}{i} \ell^{r+1-i} \right).$

Reindexing on the left and rearranging on the right yields

: $\sum_{\ell=2}^{m} \ell^{r+1} = \sum_{i=0}^{r+1} \binom{r+1}{i} \sum_{\ell=1}^{m-1} \ell^{r+1-i}$

: $\sum_{\ell=1}^{m} \ell^{r+1} = 1 + \sum_{i=0}^{r+1} \binom{r+1}{i} S_{r+1-i}(m)$

: $S_{r+1}(m+1) - S_{r+1}(m) = 1 + (r+1) S_{r}(m) + \sum_{i=2}^{r+1} \binom{r+1}{i} S_{r+1-i}(m)$

{{NumBlk|:| $m^{r+1} = 1 + (r+1) S_{r}(m) + \sum_{i=2}^{r+1} \binom{r+1}{i} S_{r+1-i}(m).$ | {{EquationRef|9}} | LnSty=0.37ex dotted Gainsboro}}

Taking {{math|1=r = k}} yields

: $m^{k+1} = 1 + (k+1) S_{k}(m) + \sum_{i=2}^{k+1} \binom{k+1}{i} S_{k+1-i}(m).$

By hypothesis, {{math|1=S_k = m^k}}, so

: $m^{k+1} = 1 + (k+1) m^k + \sum_{i=2}^{k+1} \binom{k+1}{i} S_{k+1-i}(m)$

{{NumBlk|:| $m^k ( m - (k+1) ) = 1 + \sum_{i=2}^{k+1} \binom{k+1}{i} S_{k+1-i}(m).$ | {{EquationRef|10}} | LnSty=0.37ex dotted Gainsboro}}

Since the RHS is positive, we must therefore have

{{NumBlk|:| $k + 1 < m.$ | {{EquationRef|11}} | LnSty=0.37ex dotted Gainsboro}}

Returning to ({{EquationNote|9}}) and taking {{math|1=r = k − 1}} yields

: $m^k = 1 + k \cdot S_{k-1}(m) + \sum_{i=2}^k \binom{k}{i} S_{k-i}(m)$

: $m^k = 1 + \sum_{s=1}^k \binom{k}{s} S_{k-s}(m).$

Substituting this into ({{EquationNote|10}}) to eliminate {{math|1=m^k}} yields

: $\left( 1 + \sum_{s=1}^k \binom{k}{s} S_{k-s}(m) \right) ( m - (k+1) ) = 1 + \sum_{i=2}^{k+1} \binom{k+1}{i} S_{k+1-i}(m).$

Reindexing the sum on the right with the substitution {{math|1=i = s + 1}} yields

: $\left( 1 + \sum_{s=1}^k \binom{k}{s} S_{k-s}(m) \right) ( m - (k+1) ) = 1 + \sum_{s=1}^k \binom{k+1}{s+1} S_{k-s}(m)$

: $m - (k+1) + ( m - (k+1) ) \sum_{s=1}^k \binom{k}{s} S_{k-s}(m) = 1 + \sum_{s=1}^k \frac{k+1}{s+1} \binom{k}{s} S_{k-s}(m)$

{{NumBlk|:| $m - k - 2 = \sum_{s=1}^k \left( \frac{k+1}{s+1} - m + (k+1) \right) \binom{k}{s} S_{k-s}(m).$ | {{EquationRef|12}} | LnSty=0.37ex dotted Gainsboro}}

We already know from ({{EquationNote|11}}) that {{math|1=k + 1 < m}}. This leaves open the possibility that {{math|1=m = k + 2}}; however, substituting this into ({{EquationNote|12}}) yields

: $0 = \sum_{s=1}^k \left( \frac{k+1}{s+1} - 1 \right) \binom{k}{s} S_{k-s}(k+2)$

: $0 = \sum_{s=1}^k \frac{k-s}{s+1} \binom{k}{s} S_{k-s}(k+2)$

: $0 = \frac{k-k}{k+1} \binom{k}{k} S_{k-k}(k+2) + \sum_{s=1}^{k-1} \frac{k-s}{s+1} \binom{k}{s} S_{k-s}(k+2)$

: $0 = 0 + \sum_{s=1}^{k-1} \frac{k-s}{s+1} \binom{k}{s} S_{k-s}(k+2),$

which is impossible for {{math|1=k > 1}}, since the sum contains only positive terms. Therefore any nontrivial solutions must have {{math|m ≠ k + 2}}; combining this with ({{EquationNote|11}}) yields

: $k + 2 < m.$

We therefore observe that the left-hand side of ({{EquationNote|12}}) is positive, so

{{NumBlk|:| $0 < \sum_{s=1}^k \left( \frac{k+1}{s+1} - m + (k+1) \right) \binom{k}{s} S_{k-s}(m).$ | {{EquationRef|13}} | LnSty=0.37ex dotted Gainsboro}}

Since {{math|1=k > 1}}, the sequence $\left\{ (k+1)/(s+1) - m + (k+1) \right\}_{s=1}^\infty$ is decreasing. This and ({{EquationNote|13}}) together imply that its first term (the term with {{math|1=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,

: $0 < \frac{k+1}{1+1} - m + (k+1),$

which yields

: $m < \frac{3}{2} \cdot (k+1)$

and therefore

: $\frac{m}{k+1} < \frac{3}{2},$

as desired.

Continued fractions

Any potential solutions to the equation must arise from the continued fraction of the natural logarithm of 2: specifically, {{math|1=2k / (2m − 3)}} must be a convergent of that number.

By expanding the Taylor series of {{math|1=(1 − y)^k e^ky}} about {{math|1=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

{{NumBlk|:| $\begin{align} e^{-ky} \left( 1 - \frac{k}{2} y^2 - \frac{k}{3} y^3 + \frac{k(k-2)}{8} y^4 + \frac{k(5k-6)}{30} y^5 - \frac{k^3}{6} y^6 \right) \\ < (1 - y)^k < e^{-ky} \left( 1 - \frac{k}{2} y^2 - \frac{k}{3} y^3 + \frac{k(k-2)}{8} y^4 + \frac{k^2}{2} y^5 \right) \end{align}$ |{{EquationRef|14}}|LnSty=0.37ex dotted Gainsboro}}

for {{math|1=k > 8}} and {{math|1=0 < y < 1}}.

The Erdős–Moser equation is equivalent to

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

Applying ({{EquationNote|14}}) to each term in this sum yields

: $\begin{align}
T_0 - \frac{k}{2m^2} T_2 - \frac{k}{3m^3} T_3 + \frac{k(k-2)}{8m^4} T_4 + \frac{k(5k-6)}{30m^5} T_5 - \frac{k^3}{6m^6} T_6 \qquad \qquad \\
< \sum_{j=1}^{m-1} \left(1 - \frac{j}{m} \right)^k < T_0 - \frac{k}{2m^2} T_2 - \frac{k}{3m^3} T_3 + \frac{k(k-2)}{8m^4} T_4 + \frac{k^2}{2m^5} T_5,
\end{align}$

where $T_n = \sum_{j=1}^{m-1} j^n z^j$ and {{math|1=z = e^−k/m}}. Further manipulation eventually yields

{{NumBlk|:| $\left\vert 1 - \frac{z}{1-z} + \frac{k}{2m^2}\frac{z^2+z}{(1-z)^3} + \frac{k}{3m^3}\frac{z^3+4z^2+z}{(1-z)^4} - \frac{k(k-2)}{8m^4}\frac{z^4+11z^3+11z^2+z}{(1-z)^5} \right\vert < \frac{110000}{m^3}.$ |{{EquationRef|15}}|LnSty=0.37ex dotted Gainsboro}}

We already know that {{math|1=k/m}} is bounded as {{math|1=m → ∞}}; making the ansatz {{math|1=k/m = c + O(1/m)}}, and therefore {{math|1=z = e^−c + O(1/m)}}, and substituting it into ({{EquationNote|15}}) yields

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

therefore {{math|1=c = ln(2)}}. We therefore have

{{NumBlk|:| $\frac{k}{m} = \ln(2) + \frac{a}{m} + \frac{b}{m^2} + O\left(\frac{1}{m^3}\right) \quad \text{as } m \rightarrow \infty,$ |{{EquationRef|16}}|LnSty=0.37ex dotted Gainsboro}}

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 ({{EquationNote|15}}) yields {{math|1=a = −3 ln(2) / 2}} and {{math|1=b = (3 ln(2) − 25/12) ln(2)}}. Putting these into ({{EquationNote|16}}) yields

: $\frac{k}{m} = \ln(2) \left(1 - \frac{3}{2m} - \frac{\frac{25}{12} - 3\ln(2)}{m^2} + O\left(\frac{1}{m^3}\right) \right) \quad \text{as } m \rightarrow \infty.$

The term {{Math|O(1/m³)}} must be bounded effectively. To that end, we define the function

: $F(x,\lambda) = \left.\left( 1 - \frac{1}{t-1} + \frac{x\lambda}{2} \frac{t+t^2}{(t-1)^3}
+ \frac{x^2\lambda}{3} \frac{t+4t^2+t^3}{(t-1)^4}
- \frac{x^2\lambda(\lambda-2x)}{8} \frac{t+11t^2+11t^3+t^4}{(t-1)^5}
\right) \right\vert_{t=e^\lambda}.$

The inequality ({{EquationNote|15}}) then takes the form

{{NumBlk|:| $\left\vert F \left( \frac{1}{m} , \frac{k}{m} \right) \right\vert < \frac{110000}{m^3},$ |{{EquationRef|17}}|LnSty=0.37ex dotted Gainsboro}}

and we further have

: $\begin{align}
F(x, \ln(2) (1 - \tfrac{3}{2}x \, \phantom{- \; 0.004x^2})) &> \phantom{-}0.005\phantom{15}x^2 - 100x^3 \quad \text{and} \\
F(x, \ln(2) (1 - \tfrac{3}{2}x - 0.004x^2)) &< -0.00015x^2 + 100x^3
\end{align}$

for {{Math|x ≤ 0.01}}. Therefore

: $\begin{align}
F \left( \frac{1}{m} , \ln(2) \left( 1 - \frac{3}{2m} \, \phantom{ \; - \frac{0.004}{m^2} } \right) \right) &> \phantom{-}\frac{110000}{m^3} \quad \text{for } m > 2202 \cdot 10^4 \quad \text{and} \\
F \left( \frac{1}{m} , \ln(2) \left( 1 - \frac{3}{2m} - \frac{0.004}{m^2} \right) \right) &< - \frac{110000}{m^3} \quad \text{for } m > \phantom{0}734 \cdot 10^6.
\end{align}$

Comparing these with ({{EquationNote|17}}) then shows that, for {{math|1=m > 10⁹}}, we have

: $\ln(2) \left( 1 - \frac{3}{2m} - \frac{0.004}{m^2} \right) < \frac{k}{m} < \ln(2) \left( 1 - \frac{3}{2m} \right),$

and therefore

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

Recalling that Moser showed that indeed {{math|1=m > 10⁹}}, and then invoking Legendre's theorem on continued fractions, finally proves that {{math|1=2k / (2m – 3)}} must be a convergent to {{math|1=ln(2)}}. Leveraging this result, 31 billion decimal digits of {{math|1=ln(2)}} can be used to exclude any nontrivial solutions below {{math|1=10^10⁹}}.

References

{{reflist | refs=

{{cite journal

| last1 = Butske

| first1 = William

| last2 = Jaje

| first2 = Lynda M.

| last3 = Mayernik

| first3 = Daniel R.

| year = 1999

| title = The Equation {{math|1=Σ_p|N 1/p + 1/N = 1}}, Pseudoperfect Numbers, and Partially Weighted Graphs

| journal = Mathematics of Computation

| volume = 69

| pages = 407–420

| language = English

| url = https://www.ams.org/journals/mcom/2000-69-229/S0025-5718-99-01088-1/

| access-date = 2017-03-20

| doi = 10.1090/s0025-5718-99-01088-1

| doi-access = free

| archive-url = https://web.archive.org/web/20240508231130/https://www.ams.org/journals/mcom/2000-69-229/S0025-5718-99-01088-1/S0025-5718-99-01088-1.pdf

| archive-date = 2024-05-08

}}

{{cite journal

| last1 = Gallot

| first1 = Yves

| last2 = Moree

| first2 = Pieter

| last3 = Zudilin

| first3 = Wadim

| author-link3 = Wadim Zudilin

| year = 2010

| title = The Erdős–Moser Equation {{math|1=1^k + 2^k + ⋯ + (m – 1)^k = m^k}} Revisited Using Continued Fractions

| url = https://www.ams.org/journals/mcom/2011-80-274/S0025-5718-2010-02439-1

| access-date = 2017-03-20

| journal = Mathematics of Computation

| language = English

| volume = 80

| pages = 1221–1237

| arxiv = 0907.1356

| doi = 10.1090/S0025-5718-2010-02439-1

| doi-access = free

| s2cid = 16305654

| archive-url = https://web.archive.org/web/20240508231224/https://www.ams.org/journals/mcom/2011-80-274/S0025-5718-2010-02439-1/S0025-5718-2010-02439-1.pdf

| archive-date = 2024-05-08

}}

{{cite thesis

| last = Kellner

| first = Bernd Christian

| year = 2002

| title = Über irreguläre Paare höherer Ordnungen

| url = https://www.bernoulli.org/~bk/irrpairord.pdf

| access-date = 2024-03-12

| institution = University of Göttingen

| archive-url = https://web.archive.org/web/20240312180631/https://www.bernoulli.org/~bk/irrpairord.pdf

| archive-date = 2024-03-12

| language = German

}}

{{cite journal

| last = Krzysztofek

| first = Bronisław

| year = 1966

| title = O Rówaniu {{math|1=1ⁿ + ... + mⁿ = (m + 1)ⁿ·k}}

| journal = Zeszyty Naukowe Wyżsej Szkoły Pedagogicznej w Katowicach—Sekcja Matematyki

| volume = 5

| pages = 47–54

| language = Polish

| url = https://sbc.org.pl/Content/217013/zeszyty_naukowe-sekcja_matematyki_5.pdf

| access-date = 2024-05-13

| archive-url = https://web.archive.org/web/20240513063411/https://sbc.org.pl/Content/217013/zeszyty_naukowe-sekcja_matematyki_5.pdf

| archive-date = 2024-05-13

}}

{{cite journal

| last1 = Moree

| first1 = Pieter

| last2 = te Riele

| first2 = Herman

| author-link2 = Herman te Riele

| last3 = Urbanowicz

| first3 = Jerzy

| year = 1994

| title = Divisibility Properties of Integers {{mvar|x}}, {{mvar|k}} Satisfying {{math|1=1^k + 2^k + ⋯ + (x – 1)^k = x^k}}

| url = https://www.ams.org/journals/mcom/1994-63-208/S0025-5718-1994-1257577-1/

| access-date = 2017-03-20

| journal = Mathematics of Computation

| language = English

| volume = 63

| pages = 799–815

| doi = 10.1090/s0025-5718-1994-1257577-1

| doi-access = free

| archive-url = https://web.archive.org/web/20240508230935/https://www.ams.org/journals/mcom/1994-63-208/S0025-5718-1994-1257577-1/S0025-5718-1994-1257577-1.pdf

| archive-date = 2024-05-08

}}

{{cite journal

| last = Moree

| first = Pieter

| date = 2013-10-01

| title = Moser's mathemagical work on the equation {{math|1=1^k + 2^k + ⋯ + (m – 1)^k = m^k}}

| journal = Rocky Mountain Journal of Mathematics

| volume = 43

| issue = 5

| pages = 1707–1737

| language = English

| arxiv = 1011.2940

| doi = 10.1216/RMJ-2013-43-5-1707

| doi-access = free

| issn = 0035-7596

}}

{{cite journal

| last = Moser

| first = Leo

| author-link = Leo Moser

| year = 1953

| title = On the Diophantine Equation {{math|1=1^k + 2^k + ... + (m – 1)^k = m^k}}

| journal = Scripta Mathematica

| volume = 19

| pages = 84–88

| language = English

}}

Category:Diophantine equations

Moser equation

Category:Unsolved problems in number theory

Erdős–Moser equation

Constraints on solutions

Moser's method

Bounding the ratio {{math|1=''m'' / (''k'' + 1)}}

= Method 1: Integral comparisons =

= Method 2: Algebraic manipulations =

Continued fractions

See also

References