Answer by Oscar Lanzi for 111...11 base p = 111...11 base q
Going a bit outside the box here.If we define a "Fibonacci-base" representation as$$({\ldots}a_2a_1a_0)_F=\sum_{k=1}^\infty a_kF_k$$where $a_k\in\{0,1\}$ and $F_k$ are the Fibonacci numbers, then...
View ArticleAnswer by Gerry Myerson for 111...11 base p = 111...11 base q
Richard Guy, Unsolved Problems in Number Theory, 3rd Ed., section D10, writes,The conjecture of Goormaghtigh, that the only solutions of$${x^m-1\over x-1}={y^n-1\over y-1}$$with $x,y>1$ and...
View ArticleAnswer by Dave Benson for 111...11 base p = 111...11 base q
I'm guessing that $1+5+25=31=1+2+4+8+16$ is the only example. There are certainly no more small examples, and probabilistically they get rare very quickly. But I only checked the first 80 primes to the...
View Article111...11 base p = 111...11 base q
Feels like I am probably missing something obvious.Are there distinct primes $p,q$ and positive integers $m,n$ such that$$ \sum_{i=0}^{n} p^i = \sum_{j=0}^{m} q^j$$Guessing the answer is no, but unable...
View Article