# Cyprian’s Last Theorem – proved?

To recap: we know that 3²+4²=5² and 3³+4³+5³=6³. Cyprian’s Last Theorem states that these are the only cases of n consecutive nth powers adding up to the next nth power.

I’ve nearly proved it.

Specifically, I’ve proved it for all values of n that are not of the form 16m+2 or 8m+3, and, in addition, for all values of n (of whatever form) that are less than $10^{248}$. This is not the same as proving it for absolutely all values of n, but it’s enough to be going on with.

The final gap in that proof was proving Part B of the problem described here. I floundered around for a while, getting nowhere in particular, and then submitted it to the American Mathematical Monthly for publication in its problems page. In accepting it, the editor sent me a proof by one of his colleagues of a weaker version of Part B, and I was able to strengthen it to prove the result I needed.

So now I have a 45-page paper describing the whole adventure in a chatty and discursive style: I think it’s important that mathematics should work as a spectator sport as long as the spectators can be helped to participate a little bit in what’s going on. I’m letting the paper infuse for a few weeks and then I’ll tighten it up a bit and smooth it out.

# A mathematical problem

I submitted the problem outlined here to the American Mathematical Monthly and they seem willing to publish it in their monthly problems page. One of the collaborating editors has provided a proof of part (b), which is delightful news. It takes a route I considered briefly but then discarded as being impassable!

The only thing missing in the editor’s proof is that the error in the approximation is proved to be $o(1)$, whereas I really need $O(1/n)$ and a good idea of what multiple of 1/n is involved. Perhaps one of the solvers will improve the error term once the problem appears in the magazine.

# Cyprian’s Last Theorem

I’ve been writing up my work to make it more accessible to general readers – mathematically literate but not specialists. This has expanded it from 5 pages to more than 30, has clarified and corrected some arguments, and has also led to more discoveries. In particular, the theorem now seems proved for all square-free n, and (if a particular continued-fraction argument can be made rigorous) for all n<10^76 – this is only a finite number but it’s not small.

I’m going to give the draft to some friends to read, and if it survives their scrutiny then I’ll see about posting it here, either as a single PDF or as a series of blog articles like the excellent Fermat’s Last Theorem series.

# A mathematical problem

This problem is in two parts. Part A is A-level standard; part B is more advanced.

A. Given that k > 1.0, prove that, for large n, the largest real root of $(x+1)^{n}=kx^{n}$

approaches $x\sim\frac{n}{\ln k}-\frac{1}{2}$

B. Given that k > 1.0, prove that, for large n, the largest real root of $B_{n+1}(x+1)=kB_{n+1}(x)$

approaches $x\sim\frac{n}{\ln k}+\frac{1}{k-1}+\frac{1}{2}$

# Using Bernoulli functions

Bernoulli polynomials (MathWorld, Wikipedia) are an extension of Bernoulli numbers. They obey this recurrence relation: $B_{n}(x)=(B+x)^{n}$

where $B^{n}$ is replaced by $B_{n}$ after the right-hand side has been expanded symbolically.

# Cyprian’s Last Theorem

I’ve been looking at the value N(n) of N that satisfies the equation $\sum_{i=1}^{n}(N-i)^{n}=N^{n}$

Thus turns out to be $N(n)=1.5+\frac{n}{ln2}+O(1/n)$

where the O(1/n) term is about 1/400n for n>10.

I’ve verified this by calculation up to about n=1000, using Lenstra’s long integer package LIP.

This result is so beautiful and simple that it must be possible to prove it without brute-force calculation. If anyone has any suggestions as to how to begin then I’d be very grateful.