1
$\begingroup$

Given an odd prime number $p$. Find infinitely many triples $(m,k,n)$ of integers s.t. $k,m>1 $and $k$ and $m$ are not equal, fulfilling the following relations:

$$p|2^{n+\sqrt[k]{n}}-\sqrt[m]{n}$$ and $$p\nmid 2^{n+\sqrt[m]{n}}-\sqrt[k]{n}$$

I have found the following solutions to this problem:

For $p=3$: $(m,k,n)=(5,2,(6k-4)^{10})$ $\forall k\in \mathbb N$

For $p>3$: $(m,k,n)=(4,2,[(p-1)\cdot(pk+1)-1]^8)$ $\forall k\in \mathbb N$

To check that these triples indeed are solutions is easy to see by Fermat.

How to find all solutions to the above problem?

  • 0
    $\Bbb{N}$ is infinite. You're done: +1.2012-07-07
  • 0
    Is the $\forall k\in \mathbb N$ the same as in $(m, k ,n)$ ? If not, can you change it to something else?2012-07-07
  • 0
    Yes ,what do you mean by change it to something else?2012-07-07
  • 0
    Then $\left(m=5,k=2,n=(6k-4)^{10}=(6\cdot 2-4)^{10}\right)=\left(m=5,k=2,n=8^{10}\right)$. How do you vary $k$ now? Maybe I didn't get something...2012-07-07
  • 0
    Seems like a very strange question to ask. How did this problem arise?2012-07-08

1 Answers 1