A NEW PROBLEM

[MATHPRITOM]

Let, $n>1$ be an odd number.Prove that, $n$ can not divide $3^n+1$.

[Masum]

I don't understand why people are not concious about using their common sense while posting problems. The most common problem is they don't use a good title.
Whatever, here is a solution.
Let $p$ be the smallest prime factor of $n$.
Then $3^{2n}\equiv1\pmod p$
Again from Fermat's theorem, $3^{p-1}\equiv 1\pmod p$
Then $3^{gcd(2n,p-1)}\equiv1\pmod p$
Since $p$ is the smallest prime factor, no odd primes less than $p$ can't be include in the prime factorization of $n$.
Thus, $gcd(p-1,2n)=2$.
We have, $3^2\equiv1\pmod p\Longrightarrow p=2,$ contradiction!

[tarek like math]

i think he asks is n divided by (3^n+1) i don't mention (3^n+1) is divided by n.
(3^n+1) is an even number and "n" is an odd number. generally an odd number can't divided by an even number.
if so then both "even" and "n" have a factor 2 which is not possible for a odd number "n".
sorry for mistake.

[Masum]

(3^n+1) is an even number and "n" is an odd number. generally an odd number can't divided by an even number.
if so then both "even" and "n" have a factor 2 which is not possible for a odd number "n".

Do you mean $\text{odd}\not| \text{even}$? Then what about $3|24$ and so on....