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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?

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$\Bbb{N}$ is infinite. You're done: +1. –  draks ... Jul 7 '12 at 21:06
    
Is the $\forall k\in \mathbb N$ the same as in $(m, k ,n)$ ? If not, can you change it to something else? –  draks ... Jul 7 '12 at 21:10
    
Yes ,what do you mean by change it to something else? –  Frank Jul 7 '12 at 21:36
    
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... –  draks ... Jul 7 '12 at 21:45
    
Seems like a very strange question to ask. How did this problem arise? –  Gerry Myerson Jul 8 '12 at 6:29
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1 Answer

If you rewrite $n=a^{mk}$ then the problem becomes $$ 2^{a^m(a^{m(k-1)}+1)}\equiv a^k \pmod{p} \\ 2^{a^k(a^{k(m-1)}+1)}\not\equiv a^m \pmod{p} $$ Then considering Fermat there's a good chance that $p-1\mid a^{\min(m,k)}$, $p-1\mid k$ and $p-1\nmid m$ will work provided $a\not\equiv 0,1 \pmod{p}$.

For example, for any prime with $a=p-1$, $m$ odd and $k$ a multiple of $p-1$ is a family of solutions. Or if $a=2p-2$ is a primitive root mod $p$ then any $m$ not a multiple of $p-1$ works.

There are other variants possible like your last example where $a\equiv -1 \pmod{p-1}$ and $a\equiv -2 \pmod{p}$ which generates solutions of the form $$2^{1+1}\equiv(-2)^2\not\equiv(-2)^4\pmod{p}$$ There is a family like this is taking $a\equiv -1\pmod{p-1}$ and $a\equiv 2^h \pmod{p}$ with $k=1$ and $m$ even, yielding solutions of the form $$ 2^{1+1}\equiv 2^h \not\equiv 2^{hm} \pmod{p} $$ for example if $p=11,a=59$.

Another family of possibilities is if $p-1\mid a^{m(k-1)}+1$. If we can find $x^h\equiv -1 \pmod{p-1}$ with $h>2$, then generally speaking $a\equiv x \pmod{p-1}$ and $a\equiv 1 \pmod{p}$ can work with $m=h,k=2$. For example, $p=173, a=1039, m=3, k=2$.

These are three ways to generate a large number of solutions. I may have glossed over some conditions for my constructions. But I don't have much hope that we establish precisely "all solutions," as there should be other families I have missed, and I think there may also be special cases.

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