Erdős-Moser Equation
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Theorem
The Erdős-Moser Equation is the Diophantine equation:
| \(\ds \sum_{j \mathop = 1}^k j^p\) | \(=\) | \(\ds 1 + 2^p + 3^p + \cdots + k^p\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {k + 1}^p\) |
where $k, p \in \N$.
Its only solution is $k = 2$ and $p = 1$:
- $1^1 + 2^1 = 3^1$
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Proof
Let $\map {B_n} x$ denote the $n$th Bernoulli polynomial:
- $\ds \map {B_n} x = \sum_{k \mathop = 0}^n \binom n k b_{n - k} x^k$
where $b_n$ denotes the $n$th Bernoulli number.
Lemma 1
Erdös-Moser equation is equivalent:
- $(1): \quad \ds \sum_{k \mathop = 0}^x k^p \equiv \dfrac {\map {B_{p + 1} } {x + 1} } {p + 1} = \paren {x + 1}^p$
Proof
- $\ds \sum_{k \mathop = 0}^x k^p = \dfrac {\map {B_{p + 1} } {x + 1} - \map {B_{p + 1} } 0} {p + 1}$
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It is proved that for another solution of the Erdös-Moser equation $p + 1$ must be odd and $\map {B_{p + 1} } 0$ with odd subscripts is equal to zero.
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Lemma 2
- $(2): \quad \map {B_{p + 1} } {x + 1} - \map {B_p} x = \paren {p + 1} x^p$
- $(3): \quad \map {B_{p + 1} } {x + 2} - \map {B_p} {x + 1} = \paren {p + 1} \paren {x + 1}^p$
Proof
General identity involving Bernoulli polynomials:
- $\map {B_n} {x + 1} - \map {B_n} x = n x^{n - 1}$
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Lemma 3
- $(4): \quad \dfrac {\map {B_{p + 1} } {x + 1} } {\map {B_{p + 1} } x} = \dfrac {\paren {x + 1}^p} {\paren {x + 1}^p - x^p}$
Proof
| \(\text {(5)}: \quad\) | \(\ds \dfrac {\map {B_{p + 1} } {x + 1} } {x^p} - \dfrac {\map {B_{p + 1} } x} {x^p}\) | \(=\) | \(\ds p + 1\) | expressing $p + 1$ from $(2)$ | ||||||||||
| \(\ds \map {B_{p + 1} } {x + 1}\) | \(=\) | \(\ds \paren {x + 1}^p \paren {\dfrac {\map {B_{p + 1} } {x + 1} } {x^p} - \dfrac {\map {B_{p + 1} } x} {x^p} }\) | substituting left hand side of $(5)$ for $p + 1$ in $(1)$ | |||||||||||
| \(\ds \dfrac {\map {B_{p + 1} } {x + 1} } {\map {B_{p + 1} } x}\) | \(=\) | \(\ds \dfrac {\paren {x + 1}^p} {\paren {x + 1}^p - x^p}\) | elementary rearrangements |
$\Box$
Theorem 1
The Erdös-Moser equation has other solution than $ 1 + 2 = 3$ if and only if:
- $(6): \quad \dfrac {\map {B_{p + 1} } {x + 2} } {\map {B_{p + 1} } {x + 1} } = 2$
where $x, p \in \N \land x > 2 \land p > 1$.
Proof
| \(\text {(7)}: \quad\) | \(\ds \map {B_{p + 1} } {x + 1}\) | \(=\) | \(\ds \paren {p + 1} \paren {x + 1}^p\) | rearranging $(1)$ | ||||||||||
| \(\ds \map {B_{p + 1} } {x + 2} - \map {B_p} {x + 1}\) | \(=\) | \(\ds \map {B_{p + 1} } {x + 1}\) | equating $(7)$ with $(3)$ | |||||||||||
| \(\ds \dfrac {\map {B_{p + 1} } {x + 2} } {\map {B_{p + 1} } {x + 1} }\) | \(=\) | \(\ds 2\) | elementary rearrangements |
$\Box$
Theorem 2
The Erdös-Moser equation has no solution for $p > 1$.
Proof
| \(\text {(8)}: \quad\) | \(\ds \dfrac {\map {B_{p + 1} } {x + 2} } {\map {B_{p + 1} } {x + 1} }\) | \(=\) | \(\ds \dfrac {\paren {x + 2}^p} {\paren {x + 2}^p - \paren {x + 1}^p}\) | modification of $(4)$ | ||||||||||
| \(\text {(9)}: \quad\) | \(\ds 2\) | \(=\) | \(\ds \dfrac {\paren {x + 2}^p} {\paren {x + 2}^p - \paren {x + 1}^p}\) | equating RHS of $(6)$ and RHS of $(8)$ | ||||||||||
| \(\ds 2\) | \(=\) | \(\ds \dfrac {\paren {x + 2}^p} {\paren {x + 1}^p}\) | elementary rearrangements of $(9)$ | |||||||||||
| \(\ds \sqrt [p] 2\) | \(=\) | \(\ds \dfrac {x + 2} {x + 1}\) | impossible, since $\sqrt [p] 2$ is for $p > 1$ irrational |
The sum of $p$th powers for $p = 1$:
| \(\ds \sum_{k \mathop = 0}^x k^1\) | \(=\) | \(\ds \dfrac {\map {B_2} {x + 1} - \map {B_2} 0} 2\) | ||||||||||||
| \(\text {(9)}: \quad\) | \(\ds \) | \(=\) | \(\ds \dfrac {x^2 + x} 2\) |
To prove this theorem we use $(9)$:
| \(\ds \dfrac {x^2 + x} 2\) | \(=\) | \(\ds \paren {x + 1}\) | Erdös-Moser equation, where $p = 1$ | |||||||||||
| \(\ds x^2 - x\) | \(=\) | \(\ds 2\) | elementary rearrangements | |||||||||||
| \(\ds x\) | \(=\) | \(\ds 2\) |
$\blacksquare$
Source of Name
This entry was named for Paul Erdős and Leo Moser.
Sources
- 1920: E.T. Whittaker and G.N. Watson: A Course of Modern Analysis (3rd ed.): $127$
- 1953: Leo Moser: On the diophantine equation $1^n + 2^n + 3^n + \dotsb + \paren {m − 1}^n = m^n$ (Scripta Math. Vol. 19: pp. 84 – 88)