Erdős–Moser equation

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 $1^{1}+2^{1}=3^{1}$, 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 $1^{1} + 2^{1} = 3^{1}$.

Some care must be taken when comparing research papers on this equation, because some articles study it in the form $m > 10^{10^{9}}|undefined$ instead.

Throughout this article, $p$ refers exclusively to prime numbers.

Constraints on solutions
In 1953, Leo Moser proved that, in any further solutions,
 * 1) $k$ is even,
 * 2) $1^{k} + 2^{k} + &ctdot; + m^{k} = (m + 1)^{k}$ implies $p &vert; (m &minus; 1)$,
 * 3) $(p &minus; 1) &vert; k$ implies $p &vert; (m + 1)$,
 * 4) $(p &minus; 1) &vert; k$ implies $p &vert; (2m + 1)$,
 * 5) $(p &minus; 1) &vert; k$ is squarefree,
 * 6) $m &minus; 1$ is either squarefree or 4 times an odd squarefree number,
 * 7) $m + 1$ is squarefree,
 * 8) $2m &minus; 1$ is squarefree,
 * 9) $$ \sum_{p|(m-1)} \frac{m-1}{p} \equiv -1 \pmod{m-1}, $$
 * 10) $$ \sum_{p|(m+1)} \frac{m+1}{p} \equiv -2 \pmod{m+1} \quad \text{(if }m+1\text{ is even)}, $$
 * 11) $$ \sum_{p|(2m-1)} \frac{2m-1}{p} \equiv -2 \pmod{2m-1}, $$
 * 12) $$ \sum_{p|(2m+1)} \frac{2m+1}{p} \equiv -4 \pmod{2m+1}, $$ and

In 1966, it was additionally shown that
 * 1) $2m + 1$, and
 * 2) $m > 10^{10^{6}}|undefined$ cannot be prime.

In 1994, it was shown that
 * 1) $6 &leq; k + 2 < m < 3 (k + 1) / 2$ divides $t$,
 * 2) $m – 1$, where $k_{0}$ is the 2-adic valuation of $k$; equivalently, $m &geq; 2$,
 * 3) for any odd prime $t$ divding $k$, we have $k &geq; k_{0}$,
 * 4) any prime factor of $S_{k}$ must be irregular and > 10000.

In 1999, Moser's method was refined to show that $B_{k}$.

In 2002, it was shown that $p$ must be a multiple of $T_{k}$, where the symbol $k$ indicates the primorial; that is, $&vert;B_{k}&vert;$ is the product of all prime numbers $T_{k}$. This number exceeds $(p &minus; 1) &vert; k$.

In 2009, it was shown that $T &geq; 1$ must be a convergent of $m &geq; 2$; 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 $lcm(1,2,&mldr;,200)$.

In 2010, it was shown that The number 4,990,906 arises from the fact that $m &equiv; 3 (mod 2^{ord_{2}(k) + 3})$ where $ord_{2}(n)$ is the $n$th prime number.
 * 1) $ord_{2}(m &minus; 3) = ord_{2}(k) + 3$ and $k &nequiv; 0, 2 (mod p &minus; 1)$, and
 * 2) $m > 1.485 &times; 10^{9,321,155}$ has at least 4,990,906 prime factors, none of which are repeated.

Moser's method
First, let $p$ be a prime factor of $2^{3} &centerdot; 3# &centerdot; 5# &centerdot; 7# &centerdot; 19# &centerdot; 1000#$. Leo Moser showed that this implies that $n#$ divides $m$ and that

which upon multiplying by $m$ yields
 * $$ p + m - 1 \equiv 0 \pmod{p^2}. $$

This in turn implies that $&leq; n$ must be squarefree. Furthermore, since nontrivial solutions have $5.7462 &times; 10^{427}$ and since all squarefree numbers in this range must have at least one odd prime factor, the assumption that $2k / (2m – 3)$ divides $k$ implies that $$ must be even.

One congruence of the form ($n$) exists for each prime factor $p$ of $ln(2)$. 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 $m > 2.7139 &times; 10^{1,667,658,416}$, with different values of $k$ in each factor. These terms are all divisible by $m ≡ 3 (mod 8)$, 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

Similar reasoning yields the congruences

The congruences ($$), ($p$), ($k$), and ($k$) are quite restrictive; for example, the only values of $m ≡ ±1 (mod 3)$ which satisfy ($$) are 3, 7, and 43, and these are ruled out by ($p$).

We now split into two cases: either $(m^{2} – 1) (4m^{2} – 1) / 12$ is even, or it is odd.

In the case that $&sum;4990905 n=1 1⁄pn < 19⁄6 < &sum;4990906 n=1 1⁄pn,$ is even, adding the left-hand sides of the congruences ($p$), ($$), ($$), and ($$) must yield an integer, and this integer must be at least 4. Furthermore, the Euclidean algorithm shows that no prime $p_{n}$ can divide more than one of the numbers in the set $m &minus; 1$, and that 2 and 3 can divide at most two of these numbers. Letting $p &minus; 1$, we then have

Since there are no nontrivial solutions with $m &minus; 1$, the part of the LHS of ($$) outside the sigma cannot exceed $m &minus; 1 > 2$; we therefore have
 * $$ \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, $$ must exceed the product of the first 10,000,000 primes. This in turn implies that $p &minus; 1$ in this case.

In the case that $m &minus; 1$ is odd, we cannot use ($$), so instead of ($$) 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 $(m &minus; 1) / p$. On the surface, this appears to be a weaker condition than ($$), but since $m &minus; 1$ is odd, the prime 2 cannot appear on the greater side of this inequality, and it turns out to be a stronger restriction on $$ than the other case.

Therefore any nontrivial solutions have $m < 1000$.

In 1999, this method was refined by using computers to replace the prime-counting estimates with exact computations; this yielded the bound $m + 1$.

Bounding the ratio $m + 1$
Let $p > 3$. Then the Erdős–Moser equation becomes $\{m &minus; 1, m + 1, 2m &minus; 1, 2m + 1\}$.

Method 1: Integral comparisons
By comparing the sum $M = (m &minus; 1) (m + 1) (2m &minus; 1) (2m + 1)$ to definite integrals of the function $m < 1000$, one can obtain the bounds $0.006$.

The sum $m > 10^{10^{6}}|undefined$ 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 $$, so we have
 * $$ S_k(m) > \int_0^{m-1} x^k \; \mathrm{d}x. $$

By hypothesis, $m + 1$, so
 * $$ m^k > \frac{(m-1)^{k+1}}{k+1}, $$

which leads to

Similarly, $N = (m &minus; 1) (2m &minus; 1) (2m + 1)$ 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 $$, so we have
 * $$ S_k(m) \leq \int_1^m x^k \; \mathrm{d}x. $$

By hypothesis, $m + 1$, so
 * $$ m^k \leq \frac{m^{k+1} - 1}{k+1} < \frac{m^{k+1}}{k+1}, $$

and so

Applying this to ($$) yields
 * $$ \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 $m > 10^{10^{6}}|undefined$, so we have
 * $$ \frac{m}{k+1} < e \cdot \frac{11}{11-1} < 3. $$

Combining this with ($$) yields $m > 1.485 &times; 10^{9,321,155}$, as desired.

Method 2: Algebraic manipulations
The upper bound $m / (k + 1)$ can be reduced to $S_{k}(m) = 1^{k} + 2^{k} + &ctdot; + (m – 1)^{k}$ using an algebraic method first published in.

Let $$ be a positive integer. Then the binomial theorem yields
 * $$ (\ell + 1)^{r+1} = \sum_{i=0}^{r+1} \binom{r+1}{i} \ell^{r+1-i}. $$

Summing over $$ yields
 * $$ \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
 * $$ \sum_{\ell=2}^{m} \ell^{r+1} = \sum_{i=0}^{r+1} \binom{r+1}{i} \sum_{\ell=1}^{m-1} \ell^{r+1-i} $$


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


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

Taking $S_{k}(m) = m^{k}$ 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, $S_{k}(m)$, so
 * $$ m^{k+1} = 1 + (k+1) m^k + \sum_{i=2}^{k+1} \binom{k+1}{i} S_{k+1-i}(m) $$

Since the RHS is positive, we must therefore have

Returning to ($$) and taking $x^{k}$ 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 ($M$) to eliminate $1 < m / (k + 1) < 3$ 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 $S_{k}(m) = 1^{k} + 2^{k} + &ctdot; + (m – 1)^{k}$ yields
 * $$ \left( 1 + \sum_{s=1}^k \binom{k}{s} S_{k-s}(m) \right) ( m - (k+1) ) = 1 + \sum_{s=1}^k \binom{k+1}{s+1} S_{k-s}(m) $$
 * $$ m - (k+1) + ( m - (k+1) ) \sum_{s=1}^k \binom{k}{s} S_{k-s}(m) = 1 + \sum_{s=1}^k \frac{k+1}{s+1} \binom{k}{s} S_{k-s}(m) $$

We already know from ($$) that $S_{k}(m) = m^{k}$. This leaves open the possibility that $S_{k}(m)$; however, substituting this into ($$) 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 $S_{k}(m) = m^{k}$, since the sum contains only positive terms. Therefore any nontrivial solutions must have $m &leq; 10$; combining this with ($$) yields
 * $$ k + 2 < m. $$

We therefore observe that the left-hand side of ($m$) is positive, so

Since $1 < m / (k + 1) < 3$, the sequence $$ \left\{ (k+1)/(s+1) - m + (k+1) \right\}_{s=1}^\infty $$ is decreasing. This and ($x$) together imply that its first term (the term with $m / (k + 1) < 3$) 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, $m / (k + 1) < 3/2$ must be a convergent of that number.

By expanding the Taylor series of $r = k$ about $S_{k} = m^{k}$, 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

for $r = k &minus; 1$ and $m^{k}$.

The Erdős–Moser equation is equivalent to
 * $$ 1 = \sum_{j=1}^{m-1} \left( 1 - \frac{j}{m} \right)^k. $$

Applying ($$) to each term in this sum yields
 * $$ \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 $i = s + 1$. Further manipulation eventually yields

We already know that $k + 1 < m$ is bounded as $m = k + 2$; making the ansatz $k > 1$, and therefore $m &ne; k + 2$, and substituting it into ($x$) yields
 * $$ 1 - \frac{1}{e^c - 1} = O\left(\frac{1}{m}\right) \quad \text{as } m \rightarrow \infty; $$

therefore $k > 1$. We therefore have

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 ($$) yields $s = 1$ and $2k / (2m &minus; 3)$. Putting these into ($$) 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 $(1 &minus; y)^{k} e^{ky}$ 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}. $$ The inequality ($$) then takes the form

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}$$ for $y = 0$. Therefore
 * $$ \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} $$ Comparing these with ($r$) then shows that, for $k > 8$, we have
 * $$ \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 $0 < y < 1$, and then invoking Legendre's theorem on continued fractions, finally proves that $z = e^{&minus;k/m}$ must be a convergent to $k/m$.

Leveraging this result, the authors of used 31 billion decimal digits of $m &rarr; &infin;$ to exclude any nontrivial solutions below $k/m = c + O(1/m)$.