0

Old question: The number $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$.

However, I don't understand the answer from the question. I know how to prove why it is divisible by $3^{n+1}$, but not divisible by $3^{n+2}$.

Prove it is divisible: https://math.stackexchange.com/a/1591199.

Can someone please show me step by step how to prove is not divisible?

Business Man
  • 17
  • 5
  • 1
    The duplicate says that $3\nmid (A_n-1)$, so it is not divisible by $3^{n+2}$. What is the problem? – Dietrich Burde May 09 '21 at 09:08
  • I don't understand the answer from that question, that's why I ask again. – Business Man May 09 '21 at 09:19
  • What is v3? I don't get that part – Business Man May 09 '21 at 09:36
  • An answer is also a part of this old answer of mine. Sorry to blow my own trumpet this much. – Jyrki Lahtonen May 09 '21 at 09:37
  • 1
    $\nu_3(m)$ is the exponent of the highest power of three $m$ is divisible with. So for example $\nu_3(15)=1$ and $\nu_3(162)=4$ because $3\mid 15$, $3^2\nmid 15$, $3^4\mid162$, $3^5\nmid 162$. A key property is that $\nu_3(nm)=\nu_3(n)+\nu_3(m)$ for all integers $n,m$. That, in turn, follows from the uniqueness of factorization. – Jyrki Lahtonen May 09 '21 at 09:40
  • Thanks for helping me, but I still don't quite understand it. Because I just started learning discrete maths, so I don't know any complicated symbol and stuff. Is it possible to explain an easier way to do this question? Thanks – Business Man May 09 '21 at 09:50
  • Really, Jyrki's explanation is very good and absolutely elementary. Counting how often a divisor $3$ appears, is completely elementary. If you post about divisibility by $3$, you should be willing to learn a little bit about it. – Dietrich Burde May 09 '21 at 13:28

0 Answers0