Find a positive integer $100<n<997$ such that $n|2^n+2$.
$a|b$ means $a$ divides $b$
Find an n that $n|2^n+2$
One one thing is neutral in the universe, that is $0$.
Re: Find an n that $n|2^n+2$
I couldn't find answer, but i proved rather a beautiful thing.If $n$ satisfies the condition,then $2^{n}+2$ also does!! 
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
Re: Find an n that $n|2^n+2$
Actually it is not clear. You can write it with mathematical expression.
One one thing is neutral in the universe, that is $0$.
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nafistiham
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Re: Find an n that $n|2^n+2$
Masum wrote:Actually it is not clear. You can write it with mathematical expression.
I think he wants yo say, if $n$ divides $2^n+2$
then if $m=2^n+2$
then, $m$ divides $2^m+2$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: Find an n that $n|2^n+2$
Hmm. Post the proof. Actually what we can prove is there exists infinite $n$ such that \[n|2^n+2\]
One one thing is neutral in the universe, that is $0$.
Re: Find an n that $n|2^n+2$
I think "nafistiham" has cracked the problem $99.99\%$ revealing that ,.
"I think he wants to say, if n divides $2^n+2$
then if $m=2^n+2$, then, $m$ divides $2^m+2$ "
Proof: let, $2^n+2=kn$ then $2^(2^n+2)+2= 2^{kn}+2$ which is divisible by $2^n+2$
So,this can be continued to infinity .We found a number m that satisfies the condition initially and then $k=2^m+2$ also satisfies the condition .$n$ divides $2^n+2$
Again, $p=2^k+2$ also holds the condition and thus it proceeds ....
Now, $n=2$ satisfies the condition so, $2^2+2=6$ satisfies the condition again $2^6+2$= 66 and so on.
"I think he wants to say, if n divides $2^n+2$
then if $m=2^n+2$, then, $m$ divides $2^m+2$ "
Proof: let, $2^n+2=kn$ then $2^(2^n+2)+2= 2^{kn}+2$ which is divisible by $2^n+2$
So,this can be continued to infinity .We found a number m that satisfies the condition initially and then $k=2^m+2$ also satisfies the condition .$n$ divides $2^n+2$
Again, $p=2^k+2$ also holds the condition and thus it proceeds ....
Now, $n=2$ satisfies the condition so, $2^2+2=6$ satisfies the condition again $2^6+2$= 66 and so on.
Last edited by Masum on Mon Mar 12, 2012 11:40 am, edited 2 times in total.
Reason: Please use LaTeX and proper spaces.
Reason: Please use LaTeX and proper spaces.
Re: Find an n that $n|2^n+2$
But there is a problem. The hard part is to prove that $2^{kn}+2$ is divisible by $2^n+2$. Remember that $a^n+b^n$ is divisible by $a+b$ for odd $n$.
One one thing is neutral in the universe, that is $0$.