[1] Throughout this article, p refers exclusively to prime numbers.
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.
Furthermore, since nontrivial solutions have m − 1 > 2 and since all squarefree numbers in this range must have at least one odd prime factor, the assumption that p − 1 divides k implies that k must be even.
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.
Letting M = (m − 1) (m + 1) (2m − 1) (2m + 1), we then have 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
In Moser's original paper,[2] bounds on the prime-counting function are used to observe that Therefore, M must exceed the product of the first 10,000,000 primes.
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.
By comparing the sum Sk(m) to definite integrals of the function xk, one can obtain the bounds 1 < m / (k + 1) < 3.
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 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 By hypothesis, Sk(m) = mk, so and so Applying this to (7) yields Computation shows that there are no nontrivial solutions with m ≤ 10, so we have Combining this with (8) yields 1 < m / (k + 1) < 3, as desired.
The upper bound m / (k + 1) < 3 can be reduced to m / (k + 1) < 3/2 using an algebraic method:[3]: Lemat 4 Let r be a positive integer.
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 Since k > 1, the sequence
Any potential solutions to the equation must arise from the continued fraction of the natural logarithm of 2: specifically, 2k / (2m − 3) must be a convergent of that number.
[1] By expanding the Taylor series of (1 − y)k eky about y = 0, one finds More elaborate analysis sharpens this to for k > 8 and 0 < y < 1.
The Erdős–Moser equation is equivalent to Applying (14) to each term in this sum yields where
To that end, we define the function The inequality (15) then takes the form and we further have for x ≤ 0.01.
Therefore Comparing these with (17) then shows that, for m > 109, we have and therefore Recalling that Moser showed[2] that indeed m > 109, and then invoking Legendre's theorem on continued fractions, finally proves that 2k / (2m – 3) must be a convergent to ln(2).
Leveraging this result, 31 billion decimal digits of ln(2) can be used to exclude any nontrivial solutions below 10109.