Timeline for equality of two numbers which are odd powers of 2 and satisfy a certain condition

8 events

when toggle format	what		by	license	comment
Jan 5, 2017 at 20:32	comment	added	Robert Israel		Curiously, the pair $(17,51)$ is also a counterexample to a related problem: the largest prime factor of $2^{51}+1$ divides $2^{17}+1$. This is not the case for $(37,111)$.
Jan 3, 2017 at 22:21	comment	added	Noam D. Elkies		No need for cyclotomic polynomials. In general, for any integers $b,s>1$ and $r,r'\geq 0$ we have $r \equiv r' \bmod s \Rightarrow b^r - 1 \equiv b^{r'} - 1 \bmod b^s - 1$ (using $b^s \equiv 1$). If $r'$ is the least nonnegative residue of $r \bmod s$, then $b^{r'} - 1 < b^s - 1$, so $b^r-1$ is a multiple of $b^s-1$ iff $r$ is a multiple of $s$.
Jan 3, 2017 at 21:51	comment	added	T. Amdeberhan		The equivalent condition is inherited from cyclotomic polynomials.
Jan 3, 2017 at 21:14	vote	accept	Amir Baghban
Jan 3, 2017 at 21:05	history	edited	Noam D. Elkies	CC BY-SA 3.0	added 137 characters in body
Jan 3, 2017 at 21:03	comment	added	Wojowu		I was a minute too slow, I've just found this example :P
Jan 3, 2017 at 21:03	comment	added	GH from MO		That was pretty fast!
Jan 3, 2017 at 21:03	history	answered	Noam D. Elkies	CC BY-SA 3.0