Can you help me find $n:n>41$,$\ n=41k$, and $2^x\ -\ 3^y\ =\ n$, where $k, n,x,y\in\mathbb{N}$?
I have checked all odd numbers up to $300,001$ and came up empty-handed.
Asked
Active
Viewed 46 times
1
-
2See Fermat's little theorem to get a sufficient condition on $x$ and $y$ for $n$ to be an integral multiple of $41$, i.e., of the form $n=41k$. – John Omielan Jan 12 '25 at 04:45
-
2Check again. There is a solution with both $x$ and $y$ positive (assuming you won't accept $x=y=n=0$) and at most $10$. – Jyrki Lahtonen Jan 12 '25 at 05:00
-
2Thanks. Had an error in my formula. $2^{15}-3^2,2^{35}-3^2,2^{10}-3^4$ and $2^{30}-3^4$ are small numbers that work. – Math777 Jan 12 '25 at 05:19
-
In your examples, $41$ has multiplicity $>1$. What you meant, was this post, where it is impossible. On the other hand, $41\mid 2^{20}-3^0$, and $41^2\nmid 2^{20}-3^0$, or do you define $0\not\in \Bbb N$? – Dietrich Burde Jan 12 '25 at 10:41
-
1More generally, see this post. – Dietrich Burde Jan 12 '25 at 10:49