Erdős–Moser Equation Proof (Part 22 - The Facts)

In Part 11 of this series, I said (in detail) that we are going to attempt to prove that

1k+2k+...+mk=(m+1)k1^k + 2^k + ... + m^k = (m + 1)^k

has more than 11 solution using the equation

mk(m+1)k=12km^k - (m + 1)^k = - 1 - 2^k

as

1k=11^k = 1

Now, we need to lay down the foundations.

Using the first constraint,

k=1,2k = 1, 2

as xx needs to be whole.

The second and fifth constraints would contradict the final constraint, hence no need for them.

The first half of the third constraint is useful - using OEIS, the LCM of the first 200200 natural numbers is

337293588832926264639465766794841407432394382785157234228847021917234018060677390066992000337293588832926264639465766794841407432394382785157234228847021917234018060677390066992000

The second half of the third constraint is to be used in the proof.

The fourth constraint is obsolete due to the final constraint.

As for the final constraint, we know that

m>2.73129×101,667,658,416m > 2.73129 \times 10^{1,667,658,416}

But what's the value of kk?

Going into an AMS paper referenced by the Wikipedia page of the Erdős–Moser equation (here is the paper: Erdős–Moser Equation Final Constraint Computations and Proofs Paper),

k2k \geq 2

therefore

k=2k = 2

We've got our values of m,km, k.

We're ready to do some algebraic proof!

Note by Yajat Shamji
5 months, 2 weeks ago

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