# 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.

## 2 thoughts on “Cyprian’s Last Theorem”

1. Hi!

In your equation “Sigma from 1-n (N – i)^n = N^n”

What is “i”?

Thanks for your interesting web site,

Best Wishes,

Andrew

2. “i” was the bound variable in the summation. I’ve put it in explicitly now: thank you for pointing out the deficiency!